Let it be necessary to find the numerical values of x at which several rational inequalities simultaneously turn into true numerical inequalities. In such cases, we say that we need to solve a system of rational inequalities with one unknown x.
To solve a system of rational inequalities, one must find all solutions to each inequality in the system. Then the common part of all found solutions will be the solution of the system.
Example: Solve the system of inequalities
(x -1)(x - 5)(x - 7)< 0,
First we solve the inequality
(x - 1)(x - 5)(x - 7)< 0.
Applying the interval method (Fig. 1), we find that the set of all solutions to inequality (2) consists of two intervals: (-, 1) and (5, 7).
Picture 1
Now let's solve the inequality
Using the interval method (Fig. 2), we find that the set of all solutions to inequality (3) also consists of two intervals: (2, 3) and (4, +).
Now we need to find the common part of the solution of inequalities (2) and (3). Let's draw the coordinate axis x and mark the solutions found on it. It is now clear that the common part of solving inequalities (2) and (3) is the interval (5, 7) (Fig. 3).
Consequently, the set of all solutions to the system of inequalities (1) is the interval (5, 7).
Example: Solve the system of inequalities
x2 - 6x + 10< 0,
Let's solve the inequality first
x 2 - 6x + 10< 0.
Applying the full square method, we can write that
x 2 - 6x + 10 \u003d x 2 - 2x3 + 3 2 - 3 2 + 10 \u003d (x - 3) 2 +1.
Therefore, inequality (2) can be written as
(x - 3) 2 + 1< 0,
which shows that it has no solution.
Now you can not solve the inequality
since the answer is already clear: system (1) has no solution.
Example: Solve the system of inequalities
Consider first the first inequality; we have
1 < 0, < 0.
Using the curve of signs, we find solutions to this inequality: x< -2; 0 < x < 2.
Let us now solve the second inequality of the given system. We have x 2 - 64< 0, или (х - 8)(х + 8) < 0. С помощью кривой знаков находим решения неравенства: -8 < x < 8.
Having marked the found solutions of the first and second inequalities on a common real line (Fig. 6), we find such intervals where these solutions coincide (solution suppression): -8< x < -2; 0 < x < 2. Это и есть решение системы.
Example: Solve the system of inequalities
We transform the first inequality of the system:
x 3 (x - 10) (x + 10) 0, or x (x - 10) (x + 10) 0
(since factors in odd powers can be replaced by the corresponding factors of the first degree); using the interval method, we find solutions to the last inequality: -10 x 0, x 10.
Consider the second inequality of the system; we have
We find (Fig. 8) x -9; 3< x < 15.
Combining the found solutions, we get (Fig. 9) x 0; x > 3.
Example: Find integer solutions to the system of inequalities:
x + y< 2,5,
Solution: Let's bring the system to the form
Adding the first and second inequalities, we have y< 2, 75, а учитывая третье неравенство, найдем 1 < y < 2,75. В этом интервале содержится только одно целое число 2. При y = 2 из данной системы неравенств получим
whence -1< x < 0,5. В этом интервале содержится только одно целое число 0.
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 detailed study inequalities of this kind.
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 a clear understanding of what rational inequalities are, and we 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 can 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 the 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.
Finding roots 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 third or higher 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 on the number line points 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 range of acceptable values (ODV) of the 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 not possible, 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.
Examples:
\(\frac(9x^2-1)(3x)\) \(\leq0\)
\(\frac(1)(2x)\) \(+\) \(\frac(x)(x+1)\) \(<\)\(\frac{1}{2}\)
\(\frac(6)(x+1)\) \(>\) \(\frac(x^2-5x)(x+1)\) .
When solving fractional rational inequalities, the method of intervals is used. Therefore, if the algorithm below causes you difficulties, see the article on .
How to solve fractional rational inequalities:
Algorithm for solving fractional rational inequalities.
Examples:
Place signs on the intervals of the number axis. Let me remind you the rules for arranging signs:
We determine the sign in the rightmost interval - we take a number from this interval and substitute it into the inequality instead of x. After that, we determine the signs in brackets and the result of multiplying these signs;
Examples:
Highlight the spaces you want. If there is a separate root, then mark it with a flag so that you do not forget to include it in the answer (see example below).
Examples:
Write down in response the highlighted gaps and the roots marked with a flag (if any).
Examples:
Answer: \((-∞;-1)∪(-1;1,2]∪. It consists of a set of numbers located on the coordinate line and located between -7 and 7, including borders. In this case, the points on the graph are displayed in the form of filled circles, and the interval is recorded using
The second figure is a graphical representation of the strict inequality. In this case, the boundary numbers -7 and 7, shown by punctured (not filled) dots, are not included in the specified set. And the interval itself is recorded in parentheses as follows: (-7; 7).
That is, having figured out how to solve inequalities of this type, and having received a similar answer, we can conclude that it consists of numbers that are between the considered boundaries, except for -7 and 7. The next two cases must be evaluated in a similar way. The third figure shows the images of gaps (-∞; -7] U )