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 to 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 the 9th grade, when a detailed study of inequalities of this particular type begins.
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 or more variables are practically not given any attention.
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)
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 complex 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)
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.
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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 interval method is used. Therefore, if the algorithm given below causes you difficulties, take a look at the article on .
How to solve fractional rational inequalities:
Algorithm for solving fractional rational inequalities.
Examples:
Place the signs on the number line intervals. Let me remind you of the rules for placing signs:
We determine the sign in the rightmost interval - take a number from this interval and substitute it into the inequality instead of X. After this, we determine the signs in brackets and the result of multiplying these signs;
Examples:
Select the required intervals. If there is a separate root, then mark it with a checkbox so as not to forget to include it in the answer (see example below).
Examples:
Write down the highlighted spaces and the roots marked with a flag (if any) in your answer.
Examples:
Answer: \((-∞;-1)∪(-1;1,2]∪)