The fractional inequality is not strict. Fractional rational inequalities

We continue to look at ways to solve inequalities that involve one variable. We have already studied linear and quadratic inequalities, which are special cases of rational inequalities. In this article we will clarify what type of inequalities are considered rational, and we will tell you what types they are divided into (integer and fractional). After that, we will show how to solve them correctly, provide the necessary algorithms and analyze specific problems.

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The concept of rational equalities

When they study the topic of solving inequalities in school, they immediately take rational inequalities. They acquire and hone skills in working with this type of expression. Let us formulate the definition of this concept:

Definition 1

A rational inequality is an inequality with variables that contains rational expressions in both parts.

Note that the definition does not in any way affect the question of the number of variables, which means there can be as many of them as desired. Therefore, rational inequalities with 1, 2, 3 or more variables are possible. Most often you have to deal with expressions containing only one variable, less often two, and inequalities with big amount Variables are usually not considered at all within the school course.

This way we can find out rational inequality, looking at his record. It should have rational expressions on both the right and left sides. Here are some examples:

x > 4 x 3 + 2 y ≤ 5 (y − 1) (x 2 + 1) 2 x x - 1 ≥ 1 + 1 1 + 3 x + 3 x 2

But here is an inequality of the form 5 + x + 1< x · y · z не относится к рациональным, поскольку слева у него есть переменная под знаком корня.

All rational inequalities are divided into integer and fractional.

Definition 2

The whole rational equality consists of whole rational expressions (in both parts).

Definition 3

Fractional rational equality is an equality that contains a fractional expression in one or both of its parts.

For example, inequalities of the form 1 + x - 1 1 3 2 2 + 2 3 + 2 11 - 2 1 3 x - 1 > 4 - x 4 and 1 - 2 3 5 - y > 1 x 2 - y 2 are fractional rational and 0, 5 x ≤ 3 (2 − 5 y) And 1: x + 3 > 0- whole.

We analyzed what rational inequalities are and identified their main types. We can move on to a review of ways to solve them.

Let's say that we need to find solutions to a whole rational inequality r(x)< s (x) , which includes only one variable x. Wherein r(x) And s(x) represent any integers rational numbers or expressions, and the inequality sign may differ. To solve this problem, we need to transform it and get an equivalent equality.

Let's start by moving the expression from the right side to the left. We get the following:

of the form r (x) − s (x)< 0 (≤ , > , ≥)

We know that r (x) − s (x) will be an integer value, and any integer expression can be converted to a polynomial. Let's transform r (x) − s (x) in h(x). This expression will be an identically equal polynomial. Considering that r (x) − s (x) and h (x) have the same range of permissible values ​​of x, we can move on to the inequalities h (x)< 0 (≤ , >, ≥), which will be equivalent to the original one.

Often such a simple transformation will be enough to solve the inequality, since the result may be a linear or quadratic inequality, the value of which is easy to calculate. Let's analyze such problems.

Example 1

Condition: solve a whole rational inequality x (x + 3) + 2 x ≤ (x + 1) 2 + 1.

Solution

Let's start by moving the expression from the right side to the left with the opposite sign.

x (x + 3) + 2 x − (x + 1) 2 − 1 ≤ 0

Now that we have completed all the operations with the polynomials on the left, we can move on to linear inequality 3 x − 2 ≤ 0, equivalent to what was given in the condition. It's easy to solve:

3 x ≤ 2 x ≤ 2 3

Answer: x ≤ 2 3 .

Example 2

Condition: find the solution to the inequality (x 2 + 1) 2 − 3 x 2 > (x 2 − x) (x 2 + x).

Solution

We transfer the expression from the left side to the right and perform further transformations using abbreviated multiplication formulas.

(x 2 + 1) 2 − 3 x 2 − (x 2 − x) (x 2 + x) > 0 x 4 + 2 x 2 + 1 − 3 x 2 − x 4 + x 2 > 0 1 > 0

As a result of our transformations, we received an inequality that will be true for any values ​​of x, therefore, the solution to the original inequality can be any real number.

Answer: any number really.

Example 3

Condition: solve the inequality x + 6 + 2 x 3 − 2 x (x 2 + x − 5) > 0.

Solution

We will not transfer anything from the right side, since there is 0 there. Let's start right away by converting the left side into a polynomial:

x + 6 + 2 x 3 − 2 x 3 − 2 x 2 + 10 x > 0 − 2 x 2 + 11 x + 6 > 0 .

We have derived a quadratic inequality equivalent to the original one, which can be easily solved using several methods. Let's use a graphical method.

Let's start by calculating the roots of the square trinomial − 2 x 2 + 11 x + 6:

D = 11 2 - 4 (- 2) 6 = 169 x 1 = - 11 + 169 2 - 2, x 2 = - 11 - 169 2 - 2 x 1 = - 0, 5, x 2 = 6

Now on the diagram we mark all the necessary zeros. Since the leading coefficient is less than zero, the branches of the parabola on the graph will point down.

We will need the region of the parabola located above the x-axis, since we have a > sign in the inequality. The required interval is (− 0 , 5 , 6) , therefore, this range of values ​​will be the solution we need.

Answer: (− 0 , 5 , 6) .

There are also more complex cases when a polynomial of a third or more is obtained on the left high degree. To solve such inequality, it is recommended to use the interval method. First we calculate all the roots of the polynomial h(x), which is most often done by factoring a polynomial.

Example 4

Condition: calculate (x 2 + 2) · (x + 4)< 14 − 9 · x .

Solution

Let's start, as always, by moving the expression to the left side, after which we will need to expand the brackets and bring similar terms.

(x 2 + 2) · (x + 4) − 14 + 9 · x< 0 x 3 + 4 · x 2 + 2 · x + 8 − 14 + 9 · x < 0 x 3 + 4 · x 2 + 11 · x − 6 < 0

As a result of the transformations, we got an equality equivalent to the original one, on the left of which there is a polynomial of the third degree. Let's use the interval method to solve it.

First we calculate the roots of the polynomial, for which we need to solve the cubic equation x 3 + 4 x 2 + 11 x − 6 = 0. Does it have rational roots? They can only be among the divisors of the free term, i.e. among the numbers ± 1, ± 2, ± 3, ± 6. Let's substitute them one by one into the original equation and find out that the numbers 1, 2 and 3 will be its roots.

So the polynomial x 3 + 4 x 2 + 11 x − 6 can be described as a product (x − 1) · (x − 2) · (x − 3), and inequality x 3 + 4 x 2 + 11 x − 6< 0 can be represented as (x − 1) · (x − 2) · (x − 3)< 0 . With an inequality of this type, it will then be easier for us to determine the signs on the intervals.

Next, we perform the remaining steps of the interval method: draw a number line and points on it with coordinates 1, 2, 3. They divide the straight line into 4 intervals in which they need to determine the signs. Let us shade the intervals with a minus, since the original inequality has the sign < .

All we have to do is write down the ready answer: (− ∞ , 1) ∪ (2 , 3) ​​.

Answer: (− ∞ , 1) ∪ (2 , 3) .

In some cases, proceed from the inequality r (x) − s (x)< 0 (≤ , >, ≥) to h (x)< 0 (≤ , >, ≥) , where h(x)– a polynomial to a degree higher than 2, inappropriate. This extends to cases where expressing r(x) − s(x) as a product of linear binomials and quadratic trinomials is easier than factoring h(x) into individual factors. Let's look at this problem.

Example 5

Condition: find the solution to the inequality (x 2 − 2 x − 1) (x 2 − 19) ≥ 2 x (x 2 − 2 x − 1).

Solution

This inequality applies to integers. If we move the expression from the right side to the left, open the brackets and perform a reduction of the terms, we get x 4 − 4 x 3 − 16 x 2 + 40 x + 19 ≥ 0 .

Solving such an inequality is not easy, since you have to look for the roots of a fourth-degree polynomial. It does not have a single rational root (for example, 1, − 1, 19 or − 19 are not suitable), and it is difficult to look for other roots. This means we cannot use this method.

But there are other solutions. If we move the expressions from the right side of the original inequality to the left, we can bracket the common factor x 2 − 2 x − 1:

(x 2 − 2 x − 1) (x 2 − 19) − 2 x (x 2 − 2 x − 1) ≥ 0 (x 2 − 2 x − 1) (x 2 − 2 · x − 19) ≥ 0 .

We have obtained an inequality equivalent to the original one, and its solution will give us the desired answer. Let's find the zeros of the expression on the left side, for which we solve quadratic equations x 2 − 2 x − 1 = 0 And x 2 − 2 x − 19 = 0. Their roots are 1 ± 2, 1 ± 2 5. We move on to the equality x - 1 + 2 x - 1 - 2 x - 1 + 2 5 x - 1 - 2 5 ≥ 0, which can be solved by the interval method:

According to the figure, the answer will be - ∞, 1 - 2 5 ∪ 1 - 2 5, 1 + 2 ∪ 1 + 2 5, + ∞.

Answer: - ∞ , 1 - 2 5 ∪ 1 - 2 5 , 1 + 2 ∪ 1 + 2 5 , + ∞ .

Let us add that sometimes it is not possible to find all the roots of a polynomial h(x), therefore, we cannot represent it as a product of linear binomials and quadratic trinomials. Then solve an inequality of the form h (x)< 0 (≤ , >, ≥) we cannot, which means that it is also impossible to solve the original rational inequality.

Suppose we need to solve fractionally rational inequalities of the form r (x)< s (x) (≤ , >, ≥) , where r (x) and s(x) are rational expressions, x is a variable. At least one of the indicated expressions will be fractional. The solution algorithm in this case will be as follows:

1. We determine the range of permissible values ​​of the variable x.
2. We move the expression from the right side of the inequality to the left, and the resulting expression r (x) − s (x) represent it as a fraction. Moreover, where p(x) And q(x) will be integer expressions that are products of linear binomials, indecomposable quadratic trinomials, as well as powers with a natural exponent.
3. Next, we solve the resulting inequality using the interval method.
4. The last step is to exclude the points obtained during the solution from the range of acceptable values ​​of the variable x that we defined at the beginning.

This is the algorithm for solving fractional rational inequalities. Most of it is clear; minor explanations are required only for paragraph 2. We moved the expression from the right side to the left and got r (x) − s (x)< 0 (≤ , >, ≥), and then how to bring it to the form p (x) q (x)< 0 (≤ , > , ≥) ?

First, let's determine whether this transformation can always be performed. Theoretically, such a possibility always exists, since any rational expression can be converted into a rational fraction. Here we have a fraction with polynomials in the numerator and denominator. Let us recall the fundamental theorem of algebra and Bezout's theorem and determine that any polynomial of degree n containing one variable can be transformed into a product of linear binomials. Therefore, in theory, we can always transform the expression this way.

In practice, factoring polynomials is often quite difficult, especially if the degree is higher than 4. If we cannot perform the expansion, then we will not be able to solve this inequality, but such problems are usually not studied in school courses.

Next we need to decide whether the resulting inequality p (x) q (x)< 0 (≤ , >, ≥) equivalent with respect to r (x) − s (x)< 0 (≤ , >, ≥) and to the original one. There is a possibility that it may turn out to be unequal.

The equivalence of the inequality will be ensured when the range of acceptable values p(x)q(x) will match the expression range r (x) − s (x). Then the last point of the instructions for solving fractional rational inequalities does not need to be followed.

But the range of values ​​for p(x)q(x) may be wider than r (x) − s (x), for example, by reducing fractions. An example would be going from x · x - 1 3 x - 1 2 · x + 3 to x · x - 1 x + 3 . Or this can happen when bringing similar terms, for example, here:

x + 5 x - 2 2 x - x + 5 x - 2 2 x + 1 x + 3 to 1 x + 3

For such cases, the last step of the algorithm was added. By executing it, you will get rid of extraneous variable values ​​that arise due to the expansion of the range of acceptable values. Let's take a few examples to make it more clear what we are talking about.

Example 6

Condition: find solutions to the rational equality x x + 1 · x - 3 + 4 x - 3 2 ≥ - 3 · x x - 3 2 · x + 1 .

Solution

We act according to the algorithm indicated above. First we determine the range of acceptable values. IN in this case it is determined by the system of inequalities x + 1 · x - 3 ≠ 0 x - 3 2 ≠ 0 x - 3 2 · (x + 1) ≠ 0, the solution of which is the set (− ∞, − 1) ∪ (− 1, 3) ∪ (3 , + ∞) .

x x + 1 x - 3 + 4 (x - 3) 2 + 3 x (x - 3) 2 (x + 1) ≥ 0

After that, we need to transform it so that it is convenient to apply the interval method. First of all, we reduce algebraic fractions to the lowest common denominator (x − 3) 2 (x + 1):

x x + 1 x - 3 + 4 (x - 3) 2 + 3 x (x - 3) 2 (x + 1) = = x x - 3 + 4 x + 1 + 3 x x - 3 2 x + 1 = x 2 + 4 x + 4 (x - 3) 2 (x + 1)

We collapse the expression in the numerator using the formula for the square of the sum:

x 2 + 4 x + 4 x - 3 2 x + 1 = x + 2 2 x - 3 2 x + 1

The range of acceptable values ​​of the resulting expression is (− ∞ , − 1) ∪ (− 1 , 3) ​​∪ (3 , + ∞) . We see that it is similar to what was defined for the original equality. We conclude that the inequality x + 2 2 x - 3 2 · x + 1 ≥ 0 is equivalent to the original one, which means that we do not need the last step of the algorithm.

We use the interval method:

We see the solution ( − 2 ) ∪ (− 1 , 3) ​​∪ (3 , + ∞), which will be the solution to the original rational inequality x x + 1 · x - 3 + 4 x - 3 2 ≥ - 3 · x (x - 3 ) 2 · (x + 1) .

Answer: { − 2 } ∪ (− 1 , 3) ∪ (3 , + ∞) .

Example 7

Condition: calculate the solution x + 3 x - 1 - 3 x x + 2 + 2 x - 1 > 1 x + 1 + 2 x + 2 x 2 - 1 .

Solution

We determine the range of acceptable values. In the case of this inequality, it will be equal to all real numbers except − 2, − 1, 0 and 1 .

We move the expressions from the right side to the left:

x + 3 x - 1 - 3 x x + 2 + 2 x - 1 - 1 x + 1 - 2 x + 2 x 2 - 1 > 0

x + 3 x - 1 - 3 x x + 2 = x + 3 - x - 3 x x + 2 = 0 x x + 2 = 0 x + 2 = 0

Taking into account the result, we write:

x + 3 x - 1 - 3 x x + 2 + 2 x - 1 - 1 x + 1 - 2 x + 2 x 2 - 1 = = 0 + 2 x - 1 - 1 x + 1 - 2 x + 2 x 2 - 1 = = 2 x - 1 - 1 x + 1 - 2 x + 2 x 2 - 1 = = 2 x - 1 - 1 x + 1 - 2 x + 2 (x + 1) x - 1 = = - x - 1 (x + 1) x - 1 = - x + 1 (x + 1) x - 1 = - 1 x - 1

For the expression - 1 x - 1, the range of valid values ​​is the set of all real numbers, with the exception of one. We see that the range of values ​​has expanded: − 2 , − 1 and 0 . This means we need to perform the last step of the algorithm.

Since we came to the inequality - 1 x - 1 > 0, we can write its equivalent 1 x - 1< 0 . С помощью метода интервалов вычислим решение и получим (− ∞ , 1) .

We exclude points that are not included in the range of acceptable values ​​of the original equality. We need to exclude from (− ∞ , 1) the numbers − 2 , − 1 and 0 . Thus, the solution to the rational inequality x + 3 x - 1 - 3 x x + 2 + 2 x - 1 > 1 x + 1 + 2 x + 2 x 2 - 1 will be the values ​​(− ∞ , − 2) ∪ (− 2 , − 1) ∪ (− 1 , 0) ∪ (0 , 1) .

Answer: (− ∞ , − 2) ∪ (− 2 , − 1) ∪ (− 1 , 0) ∪ (0 , 1) .

In conclusion, we give another example of a problem in which the final answer depends on the range of acceptable values.

Example 8

Condition: find the solution to the inequality 5 + 3 x 2 x 3 + 1 x 2 - x + 1 - x 2 - 1 x - 1 ≥ 0.

Solution

The range of permissible values ​​of the inequality specified in the condition is determined by the system x 2 ≠ 0 x 2 - x + 1 ≠ 0 x - 1 ≠ 0 x 3 + 1 x 2 - x + 1 - x 2 - 1 x - 1 ≠ 0.

This system has no solutions, because

x 3 + 1 x 2 - x + 1 - x 2 - 1 x - 1 = = (x + 1) x 2 - x + 1 x 2 - x + 1 - (x - 1) x + 1 x - 1 = = x + 1 - (x + 1) = 0

This means that the original equality 5 + 3 x 2 x 3 + 1 x 2 - x + 1 - x 2 - 1 x - 1 ≥ 0 has no solution, since there are no values ​​of the variable for which it would make sense.

Answer: there are no solutions.

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But today rational inequalities cannot solve everything. More precisely, not only everyone can decide. Few people can do this.
Klitschko

This lesson will be tough. So tough that only the Chosen will reach the end. Therefore, before starting reading, I recommend removing women, cats, pregnant children and... from screens.

Come on, it's actually simple. Let's say you have mastered the interval method (if you haven't mastered it, I recommend going back and reading it) and learned how to solve inequalities of the form $P\left(x \right) \gt 0$, where $P\left(x \right)$ is some polynomial or product of polynomials.

I believe that it won’t be difficult for you to solve, for example, something like this (by the way, try it as a warm-up):

\[\begin(align) & \left(2((x)^(2))+3x+4 \right)\left(4x+25 \right) \gt 0; \\ & x\left(2((x)^(2))-3x-20 \right)\left(x-1 \right)\ge 0; \\ & \left(8x-((x)^(4)) \right)((\left(x-5 \right))^(6))\le 0. \\ \end(align)\]

Now let’s complicate the problem a little and consider not just polynomials, but so-called rational fractions of the form:

where $P\left(x \right)$ and $Q\left(x \right)$ are the same polynomials of the form $((a)_(n))((x)^(n))+(( a)_(n-1))((x)^(n-1))+...+((a)_(0))$, or the product of such polynomials.

This will be a rational inequality. The fundamental point is the presence of the variable $x$ in the denominator. For example, these are rational inequalities:

\[\begin(align) & \frac(x-3)(x+7) \lt 0; \\ & \frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0; \\ & \frac(3((x)^(2))+10x+3)(((\left(3-x \right))^(2))\left(4-((x)^( 2)) \right))\ge 0. \\ \end(align)\]

And this is not a rational inequality, but the most common inequality, which can be solved by the interval method:

\[\frac(((x)^(2))+6x+9)(5)\ge 0\]

Looking ahead, I’ll say right away: there are at least two ways to solve rational inequalities, but all of them, one way or another, come down to the method of intervals already known to us. Therefore, before we analyze these methods, let's remember the old facts, otherwise there will be no sense from the new material.

What you already need to know

There are never too many important facts. We really only need four.

Abbreviated multiplication formulas

Yes, yes: they will haunt us throughout school curriculum mathematics. And at the university too. There are quite a few of these formulas, but we only need the following:

\[\begin(align) & ((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2)); \\ & ((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right); \\ & ((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b) ^(2)) \right); \\ & ((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^( 2))\right). \\ \end(align)\]

Pay attention to the last two formulas - these are the sum and difference of cubes (and not the cube of the sum or difference!). They are easy to remember if you notice that the sign in the first bracket coincides with the sign in the original expression, and in the second it is opposite to the sign in the original expression.

Linear equations

These are the most simple equations of the form $ax+b=0$, where $a$ and $b$ are ordinary numbers, and $a\ne 0$. This equation can be solved simply:

\[\begin(align) & ax+b=0; \\&ax=-b; \\ & x=-\frac(b)(a). \\ \end(align)\]

Let me note that we have the right to divide by the coefficient $a$, because $a\ne 0$. This requirement is quite logical, since for $a=0$ we get this:

First, there is no variable $x$ in this equation. This, generally speaking, should not confuse us (this happens, say, in geometry, and quite often), but still, this is no longer a linear equation.

Secondly, the solution to this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation has the form $0=0$. This equality is always true; this means $x$ is any number (usually written like this: $x\in \mathbb(R)$). If the coefficient $b$ is not equal to zero, then the equality $b=0$ is never satisfied, i.e. there are no answers (write $x\in \varnothing $ and read “the solution set is empty”).

To avoid all these difficulties, we simply assume $a\ne 0$, which does not at all limit us in further thinking.

Quadratic equations

Let me remind you that this is what a quadratic equation is called:

Here on the left is a polynomial of the second degree, and again $a\ne 0$ (otherwise, instead of quadratic equation we get linear). The following equations are solved through the discriminant:

1. If $D \gt 0$, we get two different roots;
2. If $D=0$, then the root will be the same, but of the second multiplicity (what kind of multiplicity is this and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
3. For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. By the way, this is very useful fact, which for some reason they forget to talk about in algebra lessons.

The roots themselves are calculated using the well-known formula:

\[((x)_(1,2))=\frac(-b\pm \sqrt(D))(2a)\]

Hence, by the way, the restrictions on the discriminant. After all Square root from negative number does not exist. Many students have a terrible mess in their heads about roots, so I specially wrote down a whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)

Operations with rational fractions

You already know everything that was written above if you have studied the interval method. But what we will analyze now has no analogues in the past - this is a completely new fact.

Definition. A rational fraction is an expression of the form

\[\frac(P\left(x \right))(Q\left(x \right))\]

where $P\left(x \right)$ and $Q\left(x \right)$ are polynomials.

Obviously, it’s easy to get an inequality from such a fraction—you just need to add the “greater than” or “less than” sign to the right. And a little further we will discover that solving such problems is a pleasure, everything is very simple.

Problems begin when there are several such fractions in one expression. They have to be brought to a common denominator - and it is at this moment that it is allowed a large number of offensive mistakes.

Therefore, for a successful solution rational equations Two skills need to be firmly mastered:

1. Factoring the polynomial $P\left(x \right)$;
2. Actually, bringing fractions to a common denominator.

How to factor a polynomial? Very simple. Let us have a polynomial of the form

We equate it to zero. We obtain an equation of $n$th degree:

\[((a)_(n))((x)^(n))+((a)_(n-1))((x)^(n-1))+...+(( a)_(1))x+((a)_(0))=0\]

Let's say we solved this equation and got the roots $((x)_(1)),\ ...,\ ((x)_(n))$ (don't be alarmed: in most cases there will be no more than two of these roots) . In this case, our original polynomial can be rewritten as follows:

\[\begin(align) & P\left(x \right)=((a)_(n))((x)^(n))+((a)_(n-1))((x )^(n-1))+...+((a)_(1))x+((a)_(0))= \\ & =((a)_(n))\left(x -((x)_(1)) \right)\cdot \left(x-((x)_(2)) \right)\cdot ...\cdot \left(x-((x)_( n)) \right) \end(align)\]

That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate multiplier in front of the brackets, and if necessary, it can be inserted into any of these brackets (practice shows that with $((a)_ (n))\ne \pm 1$ there are almost always fractions among the roots).

Task. Simplify the expression:

\[\frac(((x)^(2))+x-20)(x-4)-\frac(2((x)^(2))-5x+3)(2x-3)-\ frac(4-8x-5((x)^(2)))(x+2)\]

Solution. First, let's look at the denominators: they are all linear binomials, and there is nothing to factor here. So let's factor the numerators:

\[\begin(align) & ((x)^(2))+x-20=\left(x+5 \right)\left(x-4 \right); \\ & 2((x)^(2))-5x+3=2\left(x-\frac(3)(2) \right)\left(x-1 \right)=\left(2x- 3 \right)\left(x-1 \right); \\ & 4-8x-5((x)^(2))=-5\left(x+2 \right)\left(x-\frac(2)(5) \right)=\left(x +2 \right)\left(2-5x \right). \\\end(align)\]

Please note: in the second polynomial, the leading coefficient “2”, in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since the fraction appeared there.

The same thing happened in the third polynomial, only there the order of the terms is also reversed. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter the factor in one and only one bracket!), which saved us from the inconvenience associated with fractional roots.

As for the first polynomial, everything is simple: its roots are sought either standardly through the discriminant or using Vieta’s theorem.

Let's return to the original expression and rewrite it with the numerators factored:

\[\begin(matrix) \frac(\left(x+5 \right)\left(x-4 \right))(x-4)-\frac(\left(2x-3 \right)\left( x-1 \right))(2x-3)-\frac(\left(x+2 \right)\left(2-5x \right))(x+2)= \\ =\left(x+5 \right)-\left(x-1 \right)-\left(2-5x \right)= \\ =x+5-x+1-2+5x= \\ =5x+4. \\ \end(matrix)\]

Answer: $5x+4$.

As you can see, nothing complicated. A little 7th-8th grade math and that’s it. The point of all transformations is to get something simple and easy to work with from a complex and scary expression.

However, this will not always be the case. So now we will look at a more serious problem.

But first, let's figure out how to bring two fractions to a common denominator. The algorithm is extremely simple:

1. Factor both denominators;
2. Consider the first denominator and add to it factors that are present in the second denominator, but not in the first. The resulting product will be the common denominator;
3. Find out what factors each of the original fractions is missing so that the denominators become equal to the common.

This algorithm may seem to you like just text with “a lot of letters.” Therefore, let’s look at everything using a specific example.

Task. Simplify the expression:

\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]

Solution. It is better to solve such large-scale problems in parts. Let's write down what's in the first bracket:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3))-8 )-\frac(1)(x-2)\]

Unlike the previous problem, here the denominators are not so simple. Let's factor each of them.

The square trinomial $((x)^(2))+2x+4$ cannot be factorized, since the equation $((x)^(2))+2x+4=0$ has no roots (the discriminant is negative). We leave it unchanged.

The second denominator - the cubic polynomial $((x)^(3))-8$ - upon careful examination is the difference of cubes and is easily expanded using the abbreviated multiplication formulas:

\[((x)^(3))-8=((x)^(3))-((2)^(3))=\left(x-2 \right)\left(((x) ^(2))+2x+4 \right)\]

Nothing else can be factorized, since in the first bracket there is a linear binomial, and in the second there is a construction that is already familiar to us, which has no real roots.

Finally, the third denominator is a linear binomial that cannot be expanded. Thus, our equation will take the form:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))-\frac(1)(x-2)\]

It is quite obvious that the common denominator will be precisely $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$, and to reduce all fractions to it it is necessary to multiply the first fraction on $\left(x-2 \right)$, and the last one - on $\left(((x)^(2))+2x+4 \right)$. Then all that remains is to give similar ones:

\[\begin(matrix) \frac(x\cdot \left(x-2 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \ right))+\frac(((x)^(2))+8)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))- \frac(1\cdot \left(((x)^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x +4 \right))= \\ =\frac(x\cdot \left(x-2 \right)+\left(((x)^(2))+8 \right)-\left(((x )^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \\ =\frac (((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))= \\ =\frac(((x)^(2))-4x+4)(\left(x-2 \right)\ left(((x)^(2))+2x+4 \right)). \\ \end(matrix)\]

Pay attention to the second line: when the denominator is already common, i.e. Instead of three separate fractions, we wrote one big one; you shouldn’t get rid of the parentheses right away. It’s better to write an extra line and note that, say, there was a minus before the third fraction - and it won’t go anywhere, but will “hang” in the numerator in front of the bracket. This will save you from a lot of mistakes.

Well, in the last line it’s useful to factor the numerator. Moreover, this is an exact square, and abbreviated multiplication formulas again come to our aid. We have:

\[\frac(((x)^(2))-4x+4)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right) )=\frac(x-2)(((x)^(2))+2x+4)\]

Now let's deal with the second bracket in exactly the same way. Here I’ll just write a chain of equalities:

\[\begin(matrix) \frac(((x)^(2)))(((x)^(2))-4)-\frac(2)(2-x)=\frac((( x)^(2)))(\left(x-2 \right)\left(x+2 \right))-\frac(2)(2-x)= \\ =\frac(((x) ^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)= \\ =\frac(((x)^( 2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2\cdot \left(x+2 \right))(\left(x-2 \right )\cdot \left(x+2 \right))= \\ =\frac(((x)^(2))+2\cdot \left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right) ). \\ \end(matrix)\]

Let's return to the original problem and look at the product:

\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]

Answer: \[\frac(1)(x+2)\].

The meaning of this task is the same as the previous one: to show how rational expressions can be simplified if you approach their transformation wisely.

And now that you know all this, let's move on to the main topic of today's lesson - solving fractional rational inequalities. Moreover, after such preparation you will crack the inequalities themselves like nuts. :)

The main way to solve rational inequalities

There are at least two approaches to solving rational inequalities. Now we will look at one of them - the one that is generally accepted in school course mathematics.

But first, let's note an important detail. All inequalities are divided into two types:

1. Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
2. Lax: $f\left(x \right)\ge 0$ or $f\left(x \right)\le 0$.

Inequalities of the second type can easily be reduced to the first, as well as the equation:

This small “addition” $f\left(x \right)=0$ leads to such an unpleasant thing as filled points - we became familiar with them in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's look at the universal algorithm:

1. Collect all non-zero elements on one side of the inequality sign. For example, on the left;
2. Reduce all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factor the numerator and denominator. One way or another, we will get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the “tick” is the inequality sign.
3. We equate the numerator to zero: $P\left(x \right)=0$. We solve this equation and get the roots $((x)_(1))$, $((x)_(2))$, $((x)_(3))$, ... Then we require that the denominator was not equal to zero: $Q\left(x \right)\ne 0$. Of course, in essence we have to solve the equation $Q\left(x \right)=0$, and we get the roots $x_(1)^(*)$, $x_(2)^(*)$, $x_(3 )^(*)$, ... (in real problems there will hardly be more than three such roots).
4. We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punctured.
5. We place the “plus” and “minus” signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a “plus”. If $f\left(x \right) \lt 0$, then we look at the intervals with “minuses”.

Practice shows that the greatest difficulties are caused by points 2 and 4 - competent transformations and the correct arrangement of numbers in ascending order. Well, at the last step, be extremely careful: we always place signs based on the very last inequality written before moving on to the equations. This is a universal rule, inherited from the interval method.

So, there is a scheme. Let's practice.

Task. Solve the inequality:

\[\frac(x-3)(x+7) \lt 0\]

Solution. We have a strict inequality of the form $f\left(x \right) \lt 0$. Obviously, points 1 and 2 from our scheme have already been fulfilled: all the elements of inequality are collected on the left, there is no need to bring anything to a common denominator. Therefore, let's move straight to the third point.

We equate the numerator to zero:

\[\begin(align) & x-3=0; \\ & x=3. \end(align)\]

And the denominator:

\[\begin(align) & x+7=0; \\ & ((x)^(*))=-7. \\ \end(align)\]

This is where many people get stuck, because in theory you need to write $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But in the future we will be pricking out the points that came from the denominator, so there is no need to complicate your calculations again - write an equal sign everywhere and don’t worry. Nobody will deduct points for this. :)

Fourth point. We mark the resulting roots on the number line:

All points are pinned out, since the inequality is strict

Note: all points are pinned out, since the original inequality is strict. And here it doesn’t matter whether these points came from the numerator or the denominator.

Well, let's look at the signs. Let's take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but with the same success one could take $((x)_(0))=3.1$ or $((x)_(0)) =1\ 000\ 000$). We get:

So, to the right of all the roots we have a positive region. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, let’s move on to the fifth point: arrange the signs and select the one you need:

Let's return to the last inequality that was before solving the equations. Actually, it coincides with the original one, because we did not perform any transformations in this task.

Since we need to solve an inequality of the form $f\left(x \right) \lt 0$, I shaded the interval $x\in \left(-7;3 \right)$ - it is the only one marked with a minus sign. This is the answer.

Answer: $x\in \left(-7;3 \right)$

That's all! Is it difficult? No, it's not difficult. True, the task was easy. Now let’s complicate the mission a little and consider a more “sophisticated” inequality. When solving it, I will no longer give such detailed calculations - I will simply outline the key points. In general, we’ll format it the way we would format it on independent work or exam. :)

Task. Solve the inequality:

\[\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\ge 0$. All non-zero elements are collected on the left, different denominators No. Let's move on to the equations.

Numerator:

\[\begin(align) & \left(7x+1 \right)\left(11x+2 \right)=0 \\ & 7x+1=0\Rightarrow ((x)_(1))=-\ frac(1)(7); \\ & 11x+2=0\Rightarrow ((x)_(2))=-\frac(2)(11). \\ \end(align)\]

Denominator:

\[\begin(align) & 13x-4=0; \\ & 13x=4; \\ & ((x)^(*))=\frac(4)(13). \\ \end(align)\]

I don’t know what kind of pervert created this problem, but the roots didn’t turn out very well: it would be difficult to place them on the number line. And if with the root $((x)^(*))=(4)/(13)\;$ everything is more or less clear (this is the only positive number - it will be on the right), then $((x)_(1 ))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\;$ require additional research: which one is larger?

You can find this out, for example, like this:

\[((x)_(1))=-\frac(1)(7)=-\frac(2)(14) \gt -\frac(2)(11)=((x)_(2 ))\]

I hope there is no need to explain why the numerical fraction $-(2)/(14)\; \gt -(2)/(11)\;$? If necessary, I recommend remembering how to perform operations with fractions.

And we mark all three roots on the number line:

The dots from the numerator are filled in, the dots from the denominator are punctured

We are putting up signs. For example, you can take $((x)_(0))=1$ and find out the sign at this point:

\[\begin(align) & f\left(x \right)=\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4); \\ & f\left(1 \right)=\frac(\left(7\cdot 1+1 \right)\left(11\cdot 1+2 \right))(13\cdot 1-4)=\ frac(8\cdot 13)(9) \gt 0. \\\end(align)\]

The last inequality before the equations was $f\left(x \right)\ge 0$, so we are interested in the plus sign.

We got two sets: one is an ordinary segment, and the other is an open ray on the number line.

Answer: $x\in \left[ -\frac(2)(11);-\frac(1)(7) \right]\bigcup \left(\frac(4)(13);+\infty \right )$

An important note about the numbers that we substitute to find out the sign on the rightmost interval. It is absolutely not necessary to substitute the number closest to the rightmost root. You can take billions or even “plus-infinity” - in this case, the sign of the polynomial in the bracket, numerator or denominator, is determined solely by the sign of the leading coefficient.

Let's look again at the function $f\left(x \right)$ from the last inequality:

Its notation contains three polynomials:

\[\begin(align) & ((P)_(1))\left(x \right)=7x+1; \\ & ((P)_(2))\left(x \right)=11x+2; \\ & Q\left(x \right)=13x-4. \end(align)\]

All of them are linear binomials, and all of their leading coefficients (numbers 7, 11 and 13) are positive. Therefore, when substituting very large numbers, the polynomials themselves will also be positive. :)

This rule may seem overly complicated, but only at first, when we analyze very easy problems. In serious inequalities, substituting “plus-infinity” will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.

We will be faced with such challenges very soon. But first, let's look at an alternative way to solve fractional rational inequalities.

Alternative way

This technique was suggested to me by one of my students. I myself have never used it, but practice has shown that many students really find it more convenient to solve inequalities this way.

So, the initial data is the same. We need to solve the fractional rational inequality:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\]

Let's think: why is the polynomial $Q\left(x \right)$ “worse” than the polynomial $P\left(x \right)$? Why do we have to consider separate groups of roots (with and without an asterisk), think about punctured points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is different from zero.

Otherwise, there are no differences between the numerator and the denominator: we also equate it to zero, look for the roots, then mark them on the number line. So why not replace the fractional line (in fact, the division sign) with ordinary multiplication, and write down all the requirements of the ODZ in the form of a separate inequality? For example, like this:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\Rightarrow \left\( \begin(align) & P\left(x \right)\cdot Q \left(x \right) \gt 0, \\ & Q\left(x \right)\ne 0. \\ \end(align) \right.\]

Please note: this approach will reduce the problem to the interval method, but will not complicate the solution at all. After all, we will still equate the polynomial $Q\left(x \right)$ to zero.

Let's see how this works on real problems.

Task. Solve the inequality:

\[\frac(x+8)(x-11) \gt 0\]

Solution. So, let's move on to the interval method:

\[\frac(x+8)(x-11) \gt 0\Rightarrow \left\( \begin(align) & \left(x+8 \right)\left(x-11 \right) \gt 0 , \\ & x-11\ne 0. \\ \end(align) \right.\]

The first inequality can be solved in an elementary way. We simply equate each bracket to zero:

\[\begin(align) & x+8=0\Rightarrow ((x)_(1))=-8; \\ & x-11=0\Rightarrow ((x)_(2))=11. \\ \end(align)\]

The second inequality is also simple:

Mark the points $((x)_(1))$ and $((x)_(2))$ on the number line. All of them are knocked out, since the inequality is strict:

The right point was gouged out twice. This is fine.

Pay attention to the point $x=11$. It turns out that it is “double-punctured”: on the one hand, we prick it out because of the severity of inequality, on the other hand, because of the additional requirement of DL.

In any case, it will just be a punctured point. Therefore, we arrange the signs for the inequality $\left(x+8 \right)\left(x-11 \right) \gt 0$ - the last one we saw before we started solving the equations:

We are interested in positive regions, since we are solving an inequality of the form $f\left(x \right) \gt 0$ - we will shade them. All that remains is to write down the answer.

Answer. $x\in \left(-\infty ;-8 \right)\bigcup \left(11;+\infty \right)$

Using this solution as an example, I would like to warn you against a common mistake among beginning students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you from many problems.

Now let's try something more complicated.

Task. Solve the inequality:

\[\frac(\left(2x-13 \right)\left(12x-9 \right))(15x+33)\le 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\le 0$, so here you need to pay close attention to the shaded points.

Let's move on to the interval method:

\[\left\( \begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)\le 0, \\ & 15x+33\ ne 0. \\ \end(align) \right.\]

Let's go to the equation:

\[\begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0 \\ & 2x-13=0\Rightarrow ((x )_(1))=6.5; \\ & 12x-9=0\Rightarrow ((x)_(2))=0.75; \\ & 15x+33=0\Rightarrow ((x)_(3))=-2.2. \\ \end(align)\]

We take into account the additional requirement:

We mark all the resulting roots on the number line:

If a point is both punctured and filled in, it is considered to be punctured

Again, two points “overlap” each other - this is normal, it will always be like this. It is only important to understand that a point marked as both punctured and painted over is actually a punctured point. Those. "pricking" - more strong effect than "painting".

This is absolutely logical, because by pinching we mark points that affect the sign of the function, but do not themselves participate in the answer. And if at some point the number no longer suits us (for example, it does not fall into the ODZ), we cross it out from consideration until the very end of the task.

In general, stop philosophizing. We place signs and paint over those intervals that are marked with a minus sign:

Answer. $x\in \left(-\infty ;-2.2 \right)\bigcup \left[ 0.75;6.5 \right]$.

And again I wanted to draw your attention to this equation:

\[\left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0\]

Once again: never open the brackets in such equations! You will only make things more difficult for yourself. Remember: the product is equal to zero when at least one of the factors is equal to zero. Hence, given equation it simply “falls apart” into several smaller ones, which we solved in the previous problem.

Taking into account the multiplicity of roots

From the previous problems it is easy to see that it is the non-strict inequalities that are the most difficult, because in them you have to keep track of the shaded points.

But there is an even greater evil in the world - these are multiple roots in inequalities. Here you no longer have to keep track of some shaded dots - here the inequality sign may not suddenly change when passing through these same dots.

We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). Therefore, we introduce a new definition:

Definition. The root of the equation $((\left(x-a \right))^(n))=0$ is equal to $x=a$ and is called the root of the $n$th multiplicity.

Actually, we are not particularly interested exact value multiplicity. The only thing that matters is whether this same number $n$ is even or odd. Because:

1. If $x=a$ is a root of even multiplicity, then the sign of the function does not change when passing through it;
2. And vice versa, if $x=a$ is a root of odd multiplicity, then the sign of the function will change.

All previous problems discussed in this lesson are a special case of a root of odd multiplicity: everywhere the multiplicity is equal to one.

And further. Before we start solving problems, I would like to draw your attention to one subtlety that seems obvious to an experienced student, but drives many beginners into a stupor. Namely:

The root of multiplicity $n$ arises only in the case when the entire expression is raised to this power: $((\left(x-a \right))^(n))$, and not $\left(((x)^( n))-a \right)$.

Once again: the bracket $((\left(x-a \right))^(n))$ gives us the root $x=a$ of multiplicity $n$, but the bracket $\left(((x)^(n)) -a \right)$ or, as often happens, $(a-((x)^(n)))$ gives us a root (or two roots, if $n$ is even) of the first multiplicity, regardless of what equals $n$.

Compare:

\[((\left(x-3 \right))^(5))=0\Rightarrow x=3\left(5k \right)\]

Everything is clear here: the entire bracket was raised to the fifth power, so the output we got was the root of the fifth power. And now:

\[\left(((x)^(2))-4 \right)=0\Rightarrow ((x)^(2))=4\Rightarrow x=\pm 2\]

We got two roots, but both of them have first multiplicity. Or here's another one:

\[\left(((x)^(10))-1024 \right)=0\Rightarrow ((x)^(10))=1024\Rightarrow x=\pm 2\]

And don't let the tenth degree bother you. The main thing is that 10 is an even number, so at the output we have two roots, and both of them again have the first multiple.

In general, be careful: multiplicity occurs only when the degree refers to the entire parenthesis, not just the variable.

Task. Solve the inequality:

\[\frac(((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right))(((\left(x+7 \right))^(5)))\ge 0\]

Solution. Let's try to solve it in an alternative way - through the transition from the quotient to the product:

\[\left\( \begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ( (\left(x+7 \right))^(5))\ge 0, \\ & ((\left(x+7 \right))^(5))\ne 0. \\ \end(align )\right.\]

Let's deal with the first inequality using the interval method:

\[\begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ((\left( x+7 \right))^(5))=0; \\ & ((x)^(2))=0\Rightarrow x=0\left(2k \right); \\ & ((\left(6-x \right))^(3))=0\Rightarrow x=6\left(3k \right); \\ & x+4=0\Rightarrow x=-4; \\ & ((\left(x+7 \right))^(5))=0\Rightarrow x=-7\left(5k \right). \\ \end(align)\]

Additionally, we solve the second inequality. In fact, we have already solved it, but so that the reviewers do not find fault with the solution, it is better to solve it again:

\[((\left(x+7 \right))^(5))\ne 0\Rightarrow x\ne -7\]

Please note: there are no multiplicities in the last inequality. In fact: what difference does it make how many times you cross out the point $x=-7$ on the number line? At least once, at least five times, the result will be the same: a punctured point.

Let's mark everything we got on the number line:

As I said, the point $x=-7$ will eventually be punctured. The multiplicities are arranged based on solving the inequality using the interval method.

All that remains is to place the signs:

Since the point $x=0$ is a root of even multiplicity, the sign does not change when passing through it. The remaining points have an odd multiplicity, and everything is simple with them.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$

Once again, pay attention to $x=0$. Due to the even multiplicity, an interesting effect arises: everything to the left of it is painted over, everything to the right is also painted over, and the point itself is completely painted over.

As a result, it does not need to be isolated when recording the answer. Those. there is no need to write something like $x\in \left[ -4;0 \right]\bigcup \left[ 0;6 \right]$ (although formally such an answer would also be correct). Instead, we immediately write $x\in \left[ -4;6 \right]$.

Such effects are possible only with roots of even multiplicity. And in the next problem we will encounter the reverse “manifestation” of this effect. Ready?

Task. Solve the inequality:

\[\frac(((\left(x-3 \right))^(4))\left(x-4 \right))(((\left(x-1 \right))^(2)) \left(7x-10-((x)^(2)) \right))\ge 0\]

Solution. This time we'll go along standard scheme. We equate the numerator to zero:

\[\begin(align) & ((\left(x-3 \right))^(4))\left(x-4 \right)=0; \\ & ((\left(x-3 \right))^(4))=0\Rightarrow ((x)_(1))=3\left(4k \right); \\ & x-4=0\Rightarrow ((x)_(2))=4. \\ \end(align)\]

And the denominator:

\[\begin(align) & ((\left(x-1 \right))^(2))\left(7x-10-((x)^(2)) \right)=0; \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(1)^(*)=1\left(2k \right); \\ & 7x-10-((x)^(2))=0\Rightarrow x_(2)^(*)=5;\ x_(3)^(*)=2. \\ \end(align)\]

Since we are solving a non-strict inequality of the form $f\left(x \right)\ge 0$, the roots from the denominator (which have asterisks) will be taken out, and those from the numerator will be shaded.

We place signs and shade the areas marked with a “plus”:

Point $x=3$ is isolated. This is part of the answer

Before writing down the final answer, let's take a close look at the picture:

1. The point $x=1$ has an even multiplicity, but is itself punctured. Consequently, it will have to be isolated in the answer: you need to write $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left(-\ infty ;2 \right)$.
2. The point $x=3$ also has an even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step left or right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written in the form $x\in \left\( 3 \right\)$.

We combine all the received pieces into a common set and write down the answer.

Answer: $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;5 \right) $

Definition. Solving inequality means find the set of all its solutions, or prove that this set is empty.

It would seem: what could be incomprehensible here? Yes, the fact of the matter is that sets can be defined in different ways. Let's write down the answer to the last problem again:

We literally read what is written. The variable “x” belongs to a certain set, which is obtained by combining (the “U” sign) four separate sets:

• Interval $\left(-\infty ;1 \right)$, which literally means “all numbers smaller than one, but not the unit itself”;
• Interval $\left(1;2 \right)$, i.e. “all numbers in the range from 1 to 2, but not the numbers 1 and 2 themselves”;
• The set $\left\( 3 \right\)$, consisting of one single number - three;
• The interval $\left[ 4;5 \right)$ containing all numbers in the range from 4 to 5, as well as the four itself, but not the five.

The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only indicate the boundaries of these sets, the set $\left\( 3 \right\)$ specifies strictly one number by enumeration.

To understand that we are listing specific numbers included in the set (and not setting boundaries or anything else), curly braces are used. For example, the notation $\left\( 1;2 \right\)$ means exactly “a set consisting of two numbers: 1 and 2,” but not a segment from 1 to 2. Do not confuse these concepts under any circumstances.

Rule for adding multiples

Well, at the end of today's lesson, a little tin from Pavel Berdov. :)

Attentive students have probably already wondered: what will happen if the numerator and denominator have the same roots? So, the following rule works:

The multiplicities of identical roots are added. Always. Even if this root occurs in both the numerator and the denominator.

Sometimes it's better to decide than to talk. Therefore, we solve the following problem:

Task. Solve the inequality:

\[\frac(((x)^(2))+6x+8)(\left(((x)^(2))-16 \right)\left(((x)^(2))+ 9x+14 \right))\ge 0\]

\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -4. \\ \end(align)\]

Nothing special yet. We equate the denominator to zero:

\[\begin(align) & \left(((x)^(2))-16 \right)\left(((x)^(2))+9x+14 \right)=0 \\ & ( (x)^(2))-16=0\Rightarrow x_(1)^(*)=4;\ x_(2)^(*)=-4; \\ & ((x)^(2))+9x+14=0\Rightarrow x_(3)^(*)=-7;\ x_(4)^(*)=-2. \\ \end(align)\]

Two identical roots were discovered: $((x)_(1))=-2$ and $x_(4)^(*)=-2$. Both have the first multiplicity. Therefore, we replace them with one root $x_(4)^(*)=-2$, but with a multiplicity of 1+1=2.

In addition, there are also identical roots: $((x)_(2))=-4$ and $x_(2)^(*)=-4$. They are also of the first multiplicity, so only $x_(2)^(*)=-4$ of multiplicity 1+1=2 will remain.

Please note: in both cases, we left exactly the “punctured” root, and excluded the “painted” one from consideration. Because at the beginning of the lesson we agreed: if a point is both punctured and painted over, then we still consider it to be punctured.

As a result, we have four roots, and all of them were cut out:

\[\begin(align) & x_(1)^(*)=4; \\ & x_(2)^(*)=-4\left(2k \right); \\ & x_(3)^(*)=-7; \\ & x_(4)^(*)=-2\left(2k \right). \\ \end(align)\]

We mark them on the number line, taking into account the multiplicity:

We place signs and paint over the areas of interest to us:

All. No isolated points or other perversions. You can write down the answer.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left(4;+\infty \right)$.

Rule for multiplying multiples

Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to some power. In this case, the multiplicities of all original roots change.

This is rare, so most students have no experience solving such problems. And the rule here is:

When an equation is raised to the $n$ power, the multiplicities of all its roots also increase by $n$ times.

In other words, raising to a power leads to multiplying the multiples by the same power. Let's look at this rule using an example:

Task. Solve the inequality:

\[\frac(x((\left(((x)^(2))-6x+9 \right))^(2))((\left(x-4 \right))^(5)) )(((\left(2-x \right))^(3))((\left(x-1 \right))^(2)))\le 0\]

Solution. We equate the numerator to zero:

The product is zero when at least one of the factors is zero. Everything is clear with the first factor: $x=0$. But then the problems begin:

\[\begin(align) & ((\left(((x)^(2))-6x+9 \right))^(2))=0; \\ & ((x)^(2))-6x+9=0\left(2k \right); \\ & D=((6)^(3))-4\cdot 9=0 \\ & ((x)_(2))=3\left(2k \right)\left(2k \right) \ \& ((x)_(2))=3\left(4k \right) \\ \end(align)\]

As we see, the equation $((x)^(2))-6x+9=0$ has a single root of the second multiplicity: $x=3$. This entire equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which is what we eventually wrote down.

\[((\left(x-4 \right))^(5))=0\Rightarrow x=4\left(5k \right)\]

There are no problems with the denominator either:

\[\begin(align) & ((\left(2-x \right))^(3))((\left(x-1 \right))^(2))=0; \\ & ((\left(2-x \right))^(3))=0\Rightarrow x_(1)^(*)=2\left(3k \right); \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(2)^(*)=1\left(2k \right). \\ \end(align)\]

In total, we got five dots: two punctured and three painted. There are no coinciding roots in the numerator and denominator, so we simply mark them on the number line:

We arrange the signs taking into account multiplicities and paint over the intervals that interest us:

Again one isolated point and one punctured

Due to the roots of even multiplicity, we again got a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left[ 0;2 \right)$, and also an isolated point $ x\in \left\( 3 \right\)$.

Answer. $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;+\infty \right)$

As you can see, everything is not so complicated. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the same ones that we discussed at the very beginning.

Pre-conversions

The inequalities that we will examine in this section cannot be called complex. However, unlike previous tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.

We discussed this issue in detail at the very beginning of today's lesson. If you're not sure you understand what I'm talking about, I highly recommend going back and repeating it. Because there is no point in cramming methods for solving inequalities if you “float” in converting fractions.

IN homework By the way, there will also be many similar tasks. They are placed in a separate subsection. And there you will find very non-trivial examples. But this will be in homework, and now let's look at a couple of such inequalities.

Task. Solve the inequality:

\[\frac(x)(x-1)\le \frac(x-2)(x)\]

Solution. Move everything to the left:

\[\frac(x)(x-1)-\frac(x-2)(x)\le 0\]

We bring to a common denominator, open the brackets, and bring similar terms in the numerator:

\[\begin(align) & \frac(x\cdot x)(\left(x-1 \right)\cdot x)-\frac(\left(x-2 \right)\left(x-1 \ right))(x\cdot \left(x-1 \right))\le 0; \\ & \frac(((x)^(2))-\left(((x)^(2))-2x-x+2 \right))(x\left(x-1 \right)) \le 0; \\ & \frac(((x)^(2))-((x)^(2))+3x-2)(x\left(x-1 \right))\le 0; \\ & \frac(3x-2)(x\left(x-1 \right))\le 0. \\\end(align)\]

Now we have before us a classical fractional-rational inequality, the solution of which is no longer difficult. I propose to solve it using an alternative method - through the method of intervals:

\[\begin(align) & \left(3x-2 \right)\cdot x\cdot \left(x-1 \right)=0; \\ & ((x)_(1))=\frac(2)(3);\ ((x)_(2))=0;\ ((x)_(3))=1. \\ \end(align)\]

Don't forget the constraint that comes from the denominator:

We mark all the numbers and restrictions on the number line:

All roots have first multiplicity. No problem. We simply place signs and paint over the areas we need:

This is all. You can write down the answer.

Answer. $x\in \left(-\infty ;0 \right)\bigcup \left[ (2)/(3)\;;1 \right)$.

Of course, this was a very simple example. So now let’s look at the problem more seriously. And by the way, the level of this task is quite consistent with independent and tests on this topic in 8th grade.

Task. Solve the inequality:

\[\frac(1)(((x)^(2))+8x-9)\ge \frac(1)(3((x)^(2))-5x+2)\]

Solution. Move everything to the left:

\[\frac(1)(((x)^(2))+8x-9)-\frac(1)(3((x)^(2))-5x+2)\ge 0\]

Before bringing both fractions to a common denominator, let's factorize these denominators. What if the same brackets come out? With the first denominator it is easy:

\[((x)^(2))+8x-9=\left(x-1 \right)\left(x+9 \right)\]

The second one is a little more difficult. Feel free to add a constant factor into the bracket where the fraction appears. Remember: the original polynomial had integer coefficients, so there is a good chance that the factorization will have integer coefficients (in fact, it always will, unless the discriminant is irrational).

\[\begin(align) & 3((x)^(2))-5x+2=3\left(x-1 \right)\left(x-\frac(2)(3) \right)= \\ & =\left(x-1 \right)\left(3x-2 \right) \end(align)\]

As you can see, there is a common bracket: $\left(x-1 \right)$. We return to the inequality and bring both fractions to a common denominator:

\[\begin(align) & \frac(1)(\left(x-1 \right)\left(x+9 \right))-\frac(1)(\left(x-1 \right)\ left(3x-2 \right))\ge 0; \\ & \frac(1\cdot \left(3x-2 \right)-1\cdot \left(x+9 \right))(\left(x-1 \right)\left(x+9 \right )\left(3x-2 \right))\ge 0; \\ & \frac(3x-2-x-9)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ & \frac(2x-11)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ \end(align)\]

We equate the denominator to zero:

\[\begin(align) & \left(x-1 \right)\left(x+9 \right)\left(3x-2 \right)=0; \\ & x_(1)^(*)=1;\ x_(2)^(*)=-9;\ x_(3)^(*)=\frac(2)(3) \\ \end( align)\]

No multiples or coinciding roots. We mark four numbers on the line:

We are placing signs:

We write down the answer.

Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5.5;+\infty \ right)$.

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We continue to delve into the topic of “solving inequalities with one variable.” We are already familiar with linear inequalities and quadratic inequalities. They are special cases rational inequalities, which we will now study. Let's start by finding out what type of inequalities are called rational. Next we will look at their division into whole rational and fractional rational inequalities. And after this we will study how to solve rational inequalities with one variable, write down the corresponding algorithms and consider solutions typical examples with detailed explanations.

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What are rational inequalities?

In algebra classes at school, as soon as the conversation starts about solving inequalities, we immediately encounter rational inequalities. However, at first they are not called by their name, since at this stage the types of inequalities are of little interest, and the main goal is to gain initial skills in working with inequalities. The term “rational inequality” itself is introduced later in 9th grade, when detailed study inequalities of this particular type.

Let's find out what rational inequalities are. Here's the definition:

The stated definition does not say anything about the number of variables, which means that any number of them is allowed. Depending on this, rational inequalities with one, two, etc. are distinguished. variables. By the way, the textbook gives a similar definition, but for rational inequalities with one variable. This is understandable, since the school focuses on solving inequalities with one variable (below we will also talk only about solving rational inequalities with one variable). Inequalities with two variables are considered little, and inequalities with three and a large number There is almost no attention paid to variables at all.

So, a rational inequality can be recognized by its notation; to do this, just look at the expressions on its left and right sides and make sure that they are rational expressions. These considerations allow us to give examples of rational inequalities. For example, x>4 , x 3 +2 y≤5 (y−1) (x 2 +1), are rational inequalities. And inequality is not rational, since its left side contains a variable under the root sign, and, therefore, is not a rational expression. Inequality is also not rational, since both its parts are not rational expressions.

For the convenience of further description, we introduce the division of rational inequalities into integer and fractional ones.

Definition.

We will call the rational inequality whole, if both its parts are whole rational expressions.

Definition.

Fractional rational inequality is a rational inequality, at least one part of which is a fractional expression.

So 0.5 x≤3 (2−5 y) , are integer inequalities, and 1:x+3>0 and - fractionally rational.

Now we have a clear understanding of what rational inequalities are, and we can safely begin to understand the principles of solving integer and fractional rational inequalities with one variable.

Solving entire inequalities

Let’s set ourselves a task: let’s say we need to solve a whole rational inequality with one variable x of the form r(x) , ≥), where r(x) and s(x) are some integer rational expressions. To solve it, we will use equivalent inequality transformations.

Let us move the expression from the right side to the left, which will lead us to an equivalent inequality of the form r(x)−s(x)<0 (≤, >, ≥) with a zero on the right. Obviously, the expression r(x)−s(x) formed on the left side is also an integer, and it is known that any . Having transformed the expression r(x)−s(x) into the identically equal polynomial h(x) (here we note that the expressions r(x)−s(x) and h(x) have the same variable x ), we move on to the equivalent inequality h(x)<0 (≤, >, ≥).

In the simplest cases, the transformations performed will be enough to obtain the desired solution, since they will lead us from the original whole rational inequality to an inequality that we know how to solve, for example, to a linear or quadratic one. Let's look at examples.

Example.

Find the solution to the whole rational inequality x·(x+3)+2·x≤(x+1) 2 +1.

Solution.

First we move the expression from the right side to the left: x·(x+3)+2·x−(x+1) 2 −1≤0. Having completed everything on the left side, we arrive at the linear inequality 3 x−2≤0, which is equivalent to the original integer inequality. The solution is not difficult:
3 x≤2 ,
x≤2/3.

Answer:

x≤2/3.

Example.

Solve the inequality (x 2 +1) 2 −3 x 2 >(x 2 −x) (x 2 +x).

Solution.

We start as usual by transferring the expression from the right side, and then perform transformations on the left side using:
(x 2 +1) 2 −3 x 2 −(x 2 −x) (x 2 +x)>0,
x 4 +2 x 2 +1−3 x 2 −x 4 +x 2 >0,
1>0 .

Thus, by performing equivalent transformations, we arrived at the inequality 1>0, which is true for any value of the variable x. This means that the solution to the original integer inequality is any real number.

Answer:

x - any.

Example.

Solve the inequality x+6+2 x 3 −2 x (x 2 +x−5)>0.

Solution.

There is a zero on the right side, so there is no need to move anything from it. Let's transform the whole expression on the left side into a polynomial:
x+6+2 x 3 −2 x 3 −2 x 2 +10 x>0,
−2 x 2 +11 x+6>0 .

We obtained a quadratic inequality, which is equivalent to the original inequality. We solve it using any method known to us. Let's solve the quadratic inequality graphically.

Find the roots of the quadratic trinomial −2 x 2 +11 x+6 :

We make a schematic drawing on which we mark the found zeros, and take into account that the branches of the parabola are directed downward, since the leading coefficient is negative:

Since we are solving an inequality with a > sign, we are interested in the intervals in which the parabola is located above the x-axis. This occurs on the interval (−0.5, 6), which is the desired solution.

Answer:

(−0,5, 6) .

In more difficult cases on the left side of the resulting inequality h(x)<0 (≤, >, ≥) will be a polynomial of the third or higher degree. To solve such inequalities, the interval method is suitable, in the first step of which you will need to find all the roots of the polynomial h(x), which is often done through .

Example.

Find the solution to the whole rational inequality (x 2 +2)·(x+4)<14−9·x .

Solution.

Let's move everything to the left side, after which there is:
(x 2 +2)·(x+4)−14+9·x<0 ,
x 3 +4 x 2 +2 x+8−14+9 x<0 ,
x 3 +4 x 2 +11 x−6<0 .

The manipulations performed lead us to an inequality that is equivalent to the original one. On its left side is a polynomial of the third degree. It can be solved using the interval method. To do this, first of all, you need to find the roots of the polynomial that rests on x 3 +4 x 2 +11 x−6=0. Let's find out whether it has rational roots, which can only be among the divisors of the free term, that is, among the numbers ±1, ±2, ±3, ±6. Substituting these numbers in turn instead of the variable x into the equation x 3 +4 x 2 +11 x−6=0, we find out that the roots of the equation are the numbers 1, 2 and 3. This allows us to represent the polynomial x 3 +4 x 2 +11 x−6 as a product (x−1) (x−2) (x−3) , and the inequality x 3 +4 x 2 +11 x−6<0 переписать как (x−1)·(x−2)·(x−3)<0 . Такой вид неравенства в дальнейшем позволит с меньшими усилиями определить знаки на промежутках.

And then all that remains is to carry out the standard steps of the interval method: mark on the number line the points with coordinates 1, 2 and 3, which divide this line into four intervals, determine and place the signs, draw shading over the intervals with a minus sign (since we are solving an inequality with a minus sign<) и записать ответ.

Whence we have (−∞, 1)∪(2, 3) .

Answer:

(−∞, 1)∪(2, 3) .

It should be noted that sometimes it is inappropriate from the inequality r(x)−s(x)<0 (≤, >, ≥) go to the inequality h(x)<0 (≤, >, ≥), where h(x) is a polynomial of degree higher than two. This applies to cases where it is more difficult to factor the polynomial h(x) than to represent the expression r(x)−s(x) as a product of linear binomials and quadratic trinomials, for example, by factoring out the common factor. Let's explain this with an example.

Example.

Solve the inequality (x 2 −2·x−1)·(x 2 −19)≥2·x·(x 2 −2·x−1).

Solution.

This is a whole inequality. If we move the expression from its right side to the left, then open the brackets and add similar terms, we get the inequality x 4 −4 x 3 −16 x 2 +40 x+19≥0. Solving it is very difficult, since it involves finding the roots of a fourth-degree polynomial. It is easy to verify that it does not have rational roots (they could be the numbers 1, −1, 19 or −19), but it is problematic to look for its other roots. Therefore this path is a dead end.

Let's look for other possible solutions. It is easy to see that after transferring the expression from the right side of the original integer inequality to the left, we can take the common factor x 2 −2 x−1 out of brackets:
(x 2 −2·x−1)·(x 2 −19)−2·x·(x 2 −2·x−1)≥0,
(x 2 −2·x−1)·(x 2 −2·x−19)≥0.

The transformation performed is equivalent, therefore the solution to the resulting inequality will also be a solution to the original inequality.

And now we can find the zeros of the expression located on the left side of the resulting inequality, for this we need x 2 −2·x−1=0 and x 2 −2·x−19=0. Their roots are numbers . This allows us to go to the equivalent inequality, and we can solve it using the interval method:

We write down the answer according to the drawing.

Answer:

To conclude this point, I would just like to add that it is not always possible to find all the roots of the polynomial h(x) and, as a consequence, expand it into a product of linear binomials and square trinomials. In these cases there is no way to solve the inequality h(x)<0 (≤, >, ≥), which means there is no way to find a solution to the original integer rational equation.

Solving fractional rational inequalities

Now let’s solve the following problem: let’s say we need to solve a fractional rational inequality with one variable x of the form r(x) , ≥), where r(x) and s(x) are some rational expressions, and at least one of them is fractional. Let's immediately present the algorithm for solving it, after which we will make the necessary explanations.

Algorithm for solving fractional rational inequalities with one variable r(x) , ≥):

• First you need to find the range of acceptable values ​​(APV) of the variable x for the original inequality.
• Next, you need to move the expression from the right side of the inequality to the left, and convert the expression r(x)−s(x) formed there to the form of a fraction p(x)/q(x) , where p(x) and q(x) are integers expressions that are products of linear binomials, indecomposable quadratic trinomials and their powers with a natural exponent.
• Next, we need to solve the resulting inequality using the interval method.
• Finally, from the solution obtained in the previous step, it is necessary to exclude points that are not included in the ODZ of the variable x for the original inequality, which was found in the first step.

This way the desired solution to the fractional rational inequality will be obtained.

The second step of the algorithm requires explanation. Transferring the expression from the right side of the inequality to the left gives the inequality r(x)−s(x)<0 (≤, >, ≥), which is equivalent to the original one. Everything is clear here. But questions are raised by its further transformation to the form p(x)/q(x)<0 (≤, >, ≥).

The first question is: “Is it always possible to carry it out”? Theoretically, yes. We know that anything is possible. The numerator and denominator of a rational fraction contain polynomials. And from the fundamental theorem of algebra and Bezout’s theorem it follows that any polynomial of degree n with one variable can be represented as a product of linear binomials. This explains the possibility of carrying out this transformation.

In practice, it is quite difficult to factor polynomials, and if their degree is higher than four, then it is not always possible. If factorization is impossible, then there will be no way to find a solution to the original inequality, but such cases usually do not occur in school.

Second question: “Will the inequality p(x)/q(x)<0 (≤, >, ≥) is equivalent to the inequality r(x)−s(x)<0 (≤, >, ≥), and therefore to the original”? It can be either equivalent or unequal. It is equivalent when the ODZ for the expression p(x)/q(x) coincides with the ODZ for the expression r(x)−s(x) . In this case, the last step of the algorithm will be redundant. But the ODZ for the expression p(x)/q(x) may be wider than the ODZ for the expression r(x)−s(x) . Expansion of ODZ can occur when fractions are reduced, as, for example, when moving from To . Also, the expansion of ODZ can be facilitated by bringing similar terms, as, for example, when moving from To . The last step of the algorithm is intended for this case, at which extraneous decisions arising due to the expansion of the ODZ are excluded. Let's follow this when we look at the solutions to the examples below.

>>Mathematics: Rational inequalities

A rational inequality with one variable x is an inequality of the form - rational expressions, i.e. algebraic expressions made up of numbers and the variable x using the operations of addition, subtraction, multiplication, division and raising to the natural power. Of course, a variable can be denoted by any other letter, but in mathematics the letter x is most often preferred.

When solving rational inequalities, the three rules that were formulated above in § 1 are used. With the help of these rules, a given rational inequality is usually transformed to the form / (x) > 0, where / (x) is an algebraic fraction (or polynomial). Next, decompose the numerator and denominator of the fraction f (x) into factors of the form x - a (if, of course, this is possible) and apply the interval method, which we already mentioned above (see example 3 in the previous paragraph).

Example 1. Solve the inequality (x - 1) (x + 1) (x - 2) > 0.

Solution. Consider the expression f(x) = (x-1)(x + 1)(x-2).

It turns to 0 at points 1,-1,2; Let's mark these points on the number line. The number line is divided by the indicated points into four intervals (Fig. 6), at each of which the expression f (x) retains a constant sign. To verify this, let us carry out four arguments (for each of the indicated intervals separately).

Let's take any point x from the interval (2. This point is located on the number line to the right of point -1, to the right of point 1 and to the right of point 2. This means that x > -1, x > 1, x > 2 (Fig. 7). But then x-1>0, x+1>0, x - 2 > 0, and therefore f (x) > 0 (as the product of the rational inequality of three positive numbers). So, the inequality f (x) > 0 holds over the entire interval.

Let's take any point x from the interval (1,2). This point is located on the number line to the right of point-1, to the right of point 1, but to the left of point 2. This means x > -1, x > 1, but x< 2 (рис. 8), а потому x + 1>0.x-1>0.x-2<0. Но тогда f(x) <0 (как произведение двух положительных и одного отрицательного числа). Итак, на промежутке (1,2) выполняется неравенство f (x) < 0.

Let's take any point x from the interval (-1,1). This point is located on the number line to the right of point -1, to the left of point 1 and to the left of point 2. This means x > -1, but x< 1, х <2 (рис. 9), а потому х + 1 >0, x -1<0, х - 2 < 0. Но тогда f (x) >0 (as the product of two negative and one positive number). So, on the interval (-1,1) the inequality f (x)> 0 holds.

Finally, take any point x from the open ray (-oo, -1). This point is located on the number line to the left of point -1, to the left of point 1 and to the left of point 2. This means that x<-1, х< 1, х<2 (рис. 10). Но тогда x - 1 < 0, x + 1 < 0, х - 2 < 0, а значит, и f (x) < 0 (как произведение трех отрицательных чисел). Итак, на всем промежутке (-оо, -1) выполняется неравенство f (x) < 0.

Let's summarize. The signs of the expression f (x) in the selected intervals are as shown in Fig. 11. We are interested in those of them for which the inequality f (x) > 0 holds. Using the geometric model presented in Fig. 11, we establish that the inequality f (x) > 0 holds on the interval (-1, 1) or on the open ray
Answer: -1 < х < 1; х > 2.

Example 2. Solve inequality
Solution. As in the previous example, we will glean the necessary information from Fig. 11, but with two changes compared to example 1. Firstly, since we are interested in what values ​​of x the inequality f (x) holds< 0, нам придется выбрать промежутки Secondly, we are also satisfied with those points at which the equality f (x) = 0 holds. These are points -1, 1, 2, we will mark them in the figure with dark circles and include them in the answer. In Fig. Figure 12 presents a geometric model of the answer, from which it is easy to move on to analytical notation.
Answer:
Example 3. Solve inequality
Solution. Let us factorize the numerator and denominator of the algebraic fraction fx contained on the left side of the inequality. In the numerator we have x 2 - x = x(x - 1).

To factor the square trinomial x 2 - bx ~ 6 contained in the denominator of the fraction, we find its roots. From the equation x 2 - 5x - 6 = 0 we find x 1 = -1, x 2 = 6. This means (we used the formula for factoring a quadratic trinomial: ax 2 + bx + c = a(x - x 1 - x 2)).
Thus, we transformed the given inequality to the form

Consider the expression:

The numerator of this fraction turns to 0 at points 0 and 1, and turns to 0 at points -1 and 6. Let's mark these points on the number line (Fig. 13). The number line is divided by the indicated points into five intervals, and at each interval the expression fх) retains a constant sign. Reasoning in the same way as in Example 1, we come to the conclusion that the signs of the expression fх) in the selected intervals are as shown in Fig. 13. We are interested in where the inequality f (x) holds< 0. С помощью геометрической модели, представленной на рис. 13, устанавливаем, что f (х) < 0 на интервале (-1, 0) или на интервале (1, 6).

0answer: -1

Example 4. Solve inequality

Solution. When solving rational inequalities, as a rule, they prefer to leave only the number 0 on the right side of the inequality. Therefore, we transform the inequality into the form

Further:

As experience shows, if the right side of an inequality contains only the number 0, it is more convenient to carry out reasoning when on the left side both the numerator and the denominator have a positive leading coefficient. And what do we have? In our denominator, the fractions in this sense are all is in order (the leading coefficient, i.e. the coefficient of x 2, is equal to 6 - a positive number), but not everything is in order in the numerator - the leading coefficient (the coefficient of x) is equal to -4 (negative number).By multiplying both sides of the inequality by -1 and changing the sign of the inequality to the opposite, we obtain an equivalent inequality

Let's expand the numerator and denominator algebraic fraction by multipliers. In the numerator everything is simple:
To factor the square trinomial contained in the denominator of a fraction

(we again used the formula for factoring a quadratic trinomial).
Thus, we have reduced the given inequality to the form

Consider the expression

The numerator of this fraction turns to 0 at the point and the denominator - at the points. We mark these points on the number line (Fig. 14), which is divided by the indicated points into four intervals, and at each interval the expression f (x) retains a constant sign (these signs are indicated on Fig. 14). We are interested in those intervals on which the inequality fx< 0; эти промежутки выделены штриховкой на рис. 15. По условию, нас интересуют и те точки х, в которых выполняется равенство f (х) = 0. Такая точка только одна - это точка поскольку лишь при этом значении числитель дроби f (х) обращается в нуль. Точка отмечена на рис. 15 темным кружочком. Таким образом, на рис. 15 представлена геометрическая модель решения заданного неравенства, от которой нетрудно перейти к аналитической записи.

In all the examples considered, we transformed the given inequality into an equivalent inequality of the form f (x) > 0 or f (x)<0,где
In this case, the number of factors in the numerator and denominator of the fraction can be any. Then points a, b, c, d were marked on the number line. and determined the signs of the expression f (x) on the selected intervals. We noticed that on the rightmost of the selected intervals the inequality f (x) > 0 holds, and then along the intervals the signs of the expression f (x) alternate (see Fig. 16a). It is convenient to illustrate this alternation using a wavy curve, which is drawn from right to left and from top to bottom (Fig. 166). On those intervals where this curve (sometimes called the sign curve) is located above the x-axis, the inequality f (x) > 0 holds; where this curve is located below the x-axis, the inequality f (x) is satisfied< 0.

Example 5. Solve inequality

Solution. We have

(both sides of the previous inequality were multiplied by 6).
To use the interval method, mark the points on the number line (at these points the numerator of the fraction contained on the left side of the inequality becomes zero) and points (at these points the denominator of the indicated fraction becomes zero). Usually points are marked schematically, taking into account the order in which they appear (which is to the right, which is to the left) and without particularly paying attention to respect for scale. It's clear that The situation with numbers is more complicated. The first estimate shows that both numbers are slightly larger than 2.6, from which it is impossible to conclude which of the indicated numbers is larger and which is smaller. Suppose (at random) that Then
The inequality turned out to be correct, which means our guess was confirmed: in fact
So,

Let us mark the indicated 5 points in the indicated order on the number line (Fig. 17a). Let's arrange the signs of expression
on the resulting intervals: on the rightmost one there is a + sign, and then the signs alternate (Fig. 176). Let's draw a curve of signs and highlight (by shading) those intervals on which the inequality we are interested in f (x) > 0 holds (Fig. 17c). Let us finally take into account that we are talking about a non-strict inequality f (x) > 0, which means that we are also interested in those points at which the expression f (x) becomes zero. These are the roots of the numerator of the fraction f (x), i.e. points Let's mark them in Fig. 17c in dark circles (and, of course, will be included in the answer). Now here's the rice. 17c gives a complete geometric model of solutions to a given inequality.