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 kind of inequalities are called rational. Next, we will deal with their subdivision into integer rational and fractional rational inequalities. And after that we will study how the solution of rational inequalities with one variable is carried out, write down the corresponding algorithms and consider the solutions of typical examples with detailed explanations.
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What are rational inequalities?
At school, in algebra lessons, as soon as the conversation about solving inequalities comes up, the meeting with rational inequalities immediately occurs. However, at first they are not called by their proper 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 the 9th grade, when a detailed study of inequalities of this particular type begins.
Let's find out what rational inequalities are. Here is the definition:
In the voiced definition, nothing is said 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 only talk about solving rational inequalities with one variable). Inequalities with two variables are considered little, and inequalities with three and a large number variables are practically ignored.
So, a rational inequality can be recognized by its notation, for this it is enough to 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 sign of the root, and, therefore, is not a rational expression. The inequality is also not rational, since both of its parts are not rational expressions.
For the convenience of further description, we introduce the subdivision of rational inequalities into integer and fractional ones.
Definition.
Rational inequality will be called whole, if both its parts are integer rational expressions.
Definition.
Fractionally 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 clear understanding, which are rational inequalities, and you can safely begin to deal with the principles of solving integer and fractionally rational inequalities with one variable.
Solving integer inequalities
Let's set ourselves the task: let us need to solve an integer rational inequality with one variable x of the form r(x)
We 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 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 pass to the equivalent inequality h(x)<0 (≤, >, ≥).
In the simplest cases, the transformations done will be enough to get the desired solution, since they will lead us from the original integer rational inequality to an inequality that we know how to solve, for example, to a linear or square one. Consider examples.
Example.
Find a 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 done everything on the left side, we arrive at linear inequality 3 x−2≤0 , which is equivalent to the original integer inequality. His 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 moving the expression from the right side, and then we 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
.
So, performing equivalent transformations, we came to the inequality 1>0 , which is true for any values of the variable x . And 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 zero on the right side, so nothing needs to be moved 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 have obtained a quadratic inequality, which is equivalent to the original inequality. We solve it by any method known to us. We will solve the quadratic inequality graphically.
Find the roots of the square 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 downwards, since the leading coefficient is negative: 
Since we are solving the inequality with the > sign, we are interested in the intervals on which the parabola is located above the x-axis. This takes place on the interval (−0.5, 6) , and it is the desired solution.
Answer:
(−0,5, 6) .
In more complicated cases, on the left side of the resulting inequality h(x)<0 (≤, >, ≥) will be a polynomial of the third or more high degree. To solve such inequalities, the interval method is suitable, at 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 a 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 and:
(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 performed manipulations lead us to an inequality that is equivalent to the original one. On its left side is a third-degree polynomial. It can be solved using the interval method. To do this, first of all, you need to find the roots of the polynomial, which rests on x 3 +4 x 2 +11 x−6=0. Let's find out if 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 in 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 it remains to perform the standard steps of the interval method: mark points on the number line with coordinates 1, 2 and 3, which divide this line into four intervals, determine and place signs, draw hatching over the intervals with a minus sign (since we are solving an inequality with a sign<) и записать ответ.
Whence we have (−∞, 1)∪(2, 3) .
Answer:
(−∞, 1)∪(2, 3) .
It should be noted that sometimes it is impractical from the inequality r(x) − s(x)<0 (≤, >, ≥) pass to the inequality h(x)<0 (≤, >, ≥), where h(x) is a polynomial of degree greater than two. This applies to cases where it is more difficult to factorize the polynomial h(x) than to represent the expression r(x) − s(x) as a product of linear binomials and square trinomials, for example, by bracketing 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 side, then open the brackets and bring like 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 check that it does not have rational roots (they could be the numbers 1, -1, 19 or -19), and 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 side, 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 performed transformation is equivalent, so the solution of the resulting inequality will be the solution of 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 pass to an equivalent inequality , and we can solve it by the interval method: 
According to the drawing, we write down the answer.
Answer:
In conclusion of this paragraph, I would only like to add that it is far from always possible to find all the roots of the polynomial h (x) and, as a result, 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 that there is no way to find a solution to the original whole rational equation.
Solution of fractionally rational inequalities
Now let's deal with the solution of such a problem: let it be required to solve a fractionally rational inequality with one variable x of the form r(x)
Algorithm for solving a fractionally rational inequality with one variable r(x)
- First you need to find the area allowed values(ODZ) variable x for the original inequality.
- Next, you need to transfer the expression from the right side of the inequality to the left, and transform the expression r(x) − s(x) formed there into 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 square trinomials and their powers with a natural exponent.
- Next, you need to solve the resulting inequality by the method of intervals.
- Finally, from the solution obtained at the previous step, it is necessary to exclude the points that are not included in the DPV of the variable x for the original inequality, which was found at the first step.
Thus, the desired solution of the fractionally rational inequality will be obtained.
The second step of the algorithm requires some 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 are 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 the fourth, 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 at school.
Second question: “Will the inequality p(x)/q(x)<0
(≤, >, ≥) is equivalent to the inequality r(x)−s(x)<0
(≤, >, ≥), and hence also the original”? It can be either equivalent or unequal. It is equivalent when the ODZ for the expression p(x)/q(x) is the same as the ODZ for the expression r(x)−s(x) . In this case, the last step of the algorithm will be redundant. But the DPV for the expression p(x)/q(x) may be wider than the DPV for the expression r(x)−s(x) . The expansion of the ODZ can occur when fractions are reduced, as, for example, when moving from 
to . Also, the expansion of the ODZ can be facilitated by the reduction of similar terms, as, for example, in the transition from 
to . For this case, the last step of the algorithm is intended, which eliminates extraneous solutions arising from the expansion of the ODZ. Let's keep an eye on this when we analyze below the solutions of the examples.
We continue to analyze ways to solve inequalities that have one variable in their composition. 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 rational, we will tell you what types they are divided into (integer and fractional). After that, we will show how to solve them correctly, give the necessary algorithms and analyze specific problems.
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The concept of rational equalities
When the topic of solving inequalities is studied at school, they immediately take rational inequalities. They acquire and hone the skills of 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 affect the number of variables in any way, which means that there can be an arbitrarily large number of them. 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 a large number of variables are usually within school course are not considered at all.
Thus, we can learn a rational inequality by looking at its notation. Both on the right and on the left side it should have rational expressions. 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
And 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
An integer rational equality consists of integer rational expressions (in both parts).
Definition 3
Fractionally rational equality- this 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 have analyzed what rational inequalities are and identified their main types. We can move on to an overview of how to solve them.
Suppose we need to find solutions to an integer rational inequality r(x)< s (x) , which includes only one variable x . Wherein r(x) And s(x) are any integer rational numbers or expressions, and the inequality sign may be different. To solve this task, 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 possible values of x, we can pass to the inequalities h (x)< 0 (≤ , >, ≥) , which will be equivalent to the original one.
Often such a simple transformation will be sufficient to solve the inequality, since the result may be a linear or quadratic inequality, the value of which is not difficult to calculate. Let's take a look at these issues.
Example 1
Condition: solve an integer rational inequality x (x + 3) + 2 x ≤ (x + 1) 2 + 1.
Solution
Let's start by transferring the expression from the right side to the left side with the opposite sign.
x (x + 3) + 2 x − (x + 1) 2 − 1 ≤ 0
Now that we have completed all the operations with polynomials on the left, we can move on to the linear inequality 3 x − 2 ≤ 0, equivalent to what was given in the condition. Solving it is easy:
3 x ≤ 2 x ≤ 2 3
Answer: x ≤ 2 3 .
Example 2
Condition: find a 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 side and perform further transformations using the 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 got an inequality that will be true for any values of x, therefore, any real number can be the solution to the original inequality.
Answer: any real number.
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 . 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 by several methods. Let's use the graphical method.
Let's start by calculating the roots of the square trinomial − 2 x 2 + 11 x + 6:
D \u003d 11 2 - 4 (- 2) 6 \u003d 169 x 1 \u003d - 11 + 169 2 - 2, x 2 \u003d - 11 - 169 2 - 2 x 1 \u003d - 0, 5, x 2 \u003d 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 look down.
We will need a parabola area located above the abscissa axis, since we have a > sign in the inequality. The desired interval is (− 0 , 5 , 6) , therefore, this range of values will be the solution we need.
Answer: (− 0 , 5 , 6) .
There are also more complicated cases when a polynomial of the third or higher degree is obtained on the left. To solve such an 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: compute (x 2 + 2) (x + 4)< 14 − 9 · x .
Solution
Let's start, as always, by moving the expression to the left side, after which it will be necessary to open the brackets and reduce 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. We apply 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 numbers ± 1 , ± 2 , ± 3 , ± 6 . We substitute them in turn 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 presented as (x − 1) (x − 2) (x − 3)< 0 . With an inequality of this kind, 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 it is necessary to determine the signs. We shade the gaps with a minus, since the original inequality has the sign < .
We only have to write down the ready answer: (− ∞ , 1) ∪ (2 , 3) .
Answer: (− ∞ , 1) ∪ (2 , 3) .
In some cases, perform the transition from the inequality r (x) − s (x)< 0 (≤ , >, ≥) to h (x)< 0 (≤ , >, ≥) , where h(x)– a polynomial higher than 2 is inappropriate. This extends to cases where it is easier to represent r(x) − s(x) as a product of linear binomials and square trinomials than to factor h(x) into separate factors. Let's take a look at this problem.
Example 5
Condition: find a 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 the 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 any rational root (for example, 1 , − 1 , 19 or − 19 do not fit), and it is difficult to look for other roots. So we cannot use this method.
But there are other solutions as well. If we transfer the expressions from the right side of the original inequality to the left side, then we can perform the bracketing of 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 required answer. Find the zeros of the expression on the left side, for which we solve the quadratic equations x 2 − 2 x − 1 = 0 And x 2 − 2 x − 19 = 0. Their roots are 1 ± 2 , 1 ± 2 5 . We turn 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 picture, the answer is - ∞ , 1 - 2 5 ∪ 1 - 2 5 , 1 + 2 ∪ 1 + 2 5 , + ∞ .
Answer: - ∞ , 1 - 2 5 ∪ 1 - 2 5 , 1 + 2 ∪ 1 + 2 5 , + ∞ .
We 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 square trinomials. Then solve an inequality of the form h (x)< 0 (≤ , >, ≥) we cannot, therefore, 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 specified expressions will be fractional. The solution algorithm in this case will be as follows:
- We determine the range of acceptable values for the variable x .
- We transfer the expression from the right side of the inequality to the left, and the resulting expression r(x) − s(x) represented as a fraction. Meanwhile, where p(x) And q(x) will be integer expressions that are products of linear binomials, indecomposable square trinomials, as well as powers with a natural exponent.
- Next, we solve the resulting inequality by the interval method.
- The last step is to exclude the points obtained during the solution from the range of acceptable values for the x variable that we defined at the beginning.
This is the algorithm for solving a fractionally rational inequality. Most of it is clear, small 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, we determine whether a given transformation can always be performed. Theoretically, this possibility always exists, since in rational fraction can be converted to any rational expression. Here we have a fraction with polynomials in the numerator and denominator. Recall the fundamental theorem of algebra and Bezout's theorem and determine that any polynomial of nth degree containing one variable can be transformed into a product of linear binomials. Therefore, in theory, we can always transform the expression in this way.
In practice, factoring polynomials is often quite a difficult task, 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 within the framework of the school course.
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 inequality will be ensured when the range of acceptable values p(x) q(x) matches the range of the expression r(x) − s(x). Then the last paragraph of the instructions for solving fractionally rational inequalities does not need to be followed.
But the range 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 adding 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 is added. By executing it, you will get rid of the extraneous values of the variable that arise due to the expansion of the range of valid values. Let's take a few examples to make it clearer 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 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 present algebraic fractions to the lowest common denominator (x − 3) 2 (x + 1):
xx + 1 x - 3 + 4 (x - 3) 2 + 3 x (x - 3) 2 (x + 1) = = x x - 3 + 4 x + 1 + 3 xx - 3 2 x + 1 = x 2 + 4 x + 4 (x - 3) 2 (x + 1)
We collapse the expression in the numerator by applying the formula of 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 valid values of the resulting expression is (− ∞ , − 1) ∪ (− 1 , 3) ∪ (3 , + ∞) . We see that it is similar to the one that 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 xx + 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 area of admissible 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
Given the result, we write:
x + 3 x - 1 - 3 xx + 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 will be the set of all real numbers except for one. We see that the range of values has expanded: − 2 , − 1 and 0 . So, we need to perform the last step of the algorithm.
Since we have come 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 of the rational inequality x + 3 x - 1 - 3 xx + 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 one more example of a problem in which the final answer depends on the range of admissible 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 area of admissible 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 such values of the variable for which it would make sense.
Answer: there are no solutions.
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In the article we will consider solution of inequalities. Let's talk plainly about how to build a solution to inequalities with clear examples!
Before considering the solution of inequalities with examples, let's deal with the basic concepts.
Introduction to inequalities
inequality is called an expression in which functions are connected by relation signs >, . Inequalities can be both numerical and alphabetic.
Inequalities with two relation signs are called double, with three - triple, etc. For example:
a(x) > b(x),
a(x) a(x) b(x),
a(x) b(x).
a(x) Inequalities containing the sign > or or are not strict.
Inequality solution is any value of the variable for which this inequality is true.
"Solve the inequality" means that you need to find the set of all its solutions. There are various methods for solving inequalities. For inequality solutions use a number line that is infinite. For example, solving the inequality x > 3 is an interval from 3 to +, and the number 3 is not included in this interval, so the point on the line is denoted by an empty circle, because the inequality is strict. 
+
The answer will be: x (3; +).
The value x=3 is not included in the set of solutions, so the parenthesis is round. The infinity sign is always enclosed in a parenthesis. The sign means "belonging".
Consider how to solve inequalities using another example with the sign:
x2
-
+
The value x=2 is included in the set of solutions, so the square bracket and the point on the line is denoted by a filled circle.
The answer will be: x $.
How to introduce a new variable
This method is as follows: An equation of the form $f(x)=g(x)$ is written. We solve it as follows: we introduce such a new variable in order to obtain an equation whose solution is already known. We subsequently solve it and return to the replacement. From it we find the solution of the first equation. Further, the found roots are marked on the number line and a curve of signs is constructed. Depending on the sign of the initial inequality, the answer is written.
