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 a large number of variables are usually not considered at all in the school course.
Thus, we can recognize a rational inequality by looking at its writing. 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 rational integer 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 the 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 the third or higher degree is obtained on the left. 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:
- We determine the range of permissible values of the variable x.
- 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.
- Next, we solve the resulting inequality using the interval method.
- 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 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 will be 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 except 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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Interval method– a simple way to solve fractional rational inequalities. This is the name for inequalities containing rational (or fractional-rational) expressions that depend on a variable.
1. Consider, for example, the following inequality
The interval method allows you to solve it in a couple of minutes.
On the left side of this inequality is a fractional rational function. Rational because it does not contain roots, sines, or logarithms - only rational expressions. On the right is zero.
The interval method is based on the following property of a fractional rational function.
A fractional rational function can change sign only at those points at which it is equal to zero or does not exist.
Let us recall how a quadratic trinomial is factored, that is, an expression of the form .
Where and are the roots of the quadratic equation.
We draw an axis and place the points at which the numerator and denominator go to zero.
The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded, since the inequality is not strict. When and our inequality is satisfied, since both its sides are equal to zero.
These points break the axis into intervals.
Let us determine the sign of the fractional rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points at which it is equal to zero or does not exist. This means that at each of the intervals between the points where the numerator or denominator goes to zero, the sign of the expression on the left side of the inequality will be constant - either “plus” or “minus”.
And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that is convenient for us.
. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.
Next interval: . Let's check the sign at . We find that the left side has changed its sign to .
Let's take it. When the expression is positive - therefore, it is positive over the entire interval from to.
When the left side of the inequality is negative.
And finally, class="tex" alt="x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}
We have found at what intervals the expression is positive. All that remains is to write down the answer:
Answer: .
Please note: the signs alternate between intervals. This happened because when passing through each point, exactly one of the linear factors changed sign, while the rest kept it unchanged.
We see that the interval method is very simple. To solve the fractional-rational inequality using the interval method, we reduce it to the form:
Or class="tex" alt="\genfrac())()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or or .
(on the left side is a fractional rational function, on the right side is zero).
Then we mark on the number line the points at which the numerator or denominator goes to zero.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
All that remains is to find out its sign at each interval.
We do this by checking the sign of the expression at any point belonging to a given interval. After that, we write down the answer. That's all.
But the question arises: do the signs always alternate? No not always! You must be careful and not place signs mechanically and thoughtlessly.
2. Let's consider another inequality.
Class="tex" alt="\genfrac())()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3 \right))>0"> !}
Place the points on the axis again. The dots and are punctured because they are zeros of the denominator. The point is also cut out, since the inequality is strict.
When the numerator is positive, both factors in the denominator are negative. This can be easily checked by taking any number from a given interval, for example, . The left side has the sign:
When the numerator is positive; The first factor in the denominator is positive, the second factor is negative. The left side has the sign:
The situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:
Finally, with class="tex" alt="x>3">
все множители положительны, и левая часть имеет знак :!} 
Answer: .
Why was the alternation of signs disrupted? Because when passing through a point the multiplier is “responsible” for it didn't change sign. Consequently, the entire left side of our inequality did not change sign.
Conclusion: if the linear multiplier is an even power (for example, squared), then when passing through a point the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.
3. Let's consider a more complex case. It differs from the previous one in that the inequality is not strict:
The left side is the same as in the previous problem. The picture of signs will be the same:
Maybe the answer will be the same? No! A solution is added This happens because at both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.
Answer: .
This situation often occurs in problems on the Unified State Examination in mathematics. This is where applicants fall into a trap and lose points. Be careful!
4. What to do if the numerator or denominator cannot be factored into linear factors? Consider this inequality:
A square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression for all is the same, and specifically, positive. You can read more about this in the article on properties of quadratic functions.
And now we can divide both sides of our inequality by a value that is positive for all. Let us arrive at an equivalent inequality:
Which is easily solved using the interval method.
Please note that we divided both sides of the inequality by a value that we knew for sure was positive. Of course, in general, you should not multiply or divide an inequality by a variable whose sign is unknown.
5 . Let's consider another inequality, seemingly quite simple:
I just want to multiply it by . But we are already smart, and we won’t do this. After all, it can be both positive and negative. And we know that if both sides of the inequality are multiplied by a negative value, the sign of the inequality changes.
We will do it differently - we will collect everything in one part and bring it to a common denominator. The right side will remain zero:
Class="tex" alt="\genfrac())()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}
And after that - apply interval method.
- Develop the ability to solve rational inequalities using the method of intervals with multiple roots, help students develop the need and desire to generalize the material studied;
- Develop the ability to compare solutions and identify the correct answers; develop curiosity, logical thinking, cognitive interest in the subject
- Cultivate accuracy when drawing up solutions, the ability to overcome difficulties when solving inequalities.
Materials and equipment: interactive whiteboard, cards, collection of tests.
Progress of the lesson
I. Organizational moment
II. Updating knowledge
Frontal class survey on the following questions:
At what values of the variable does the fraction make sense (Fig. 1)?
Repeat the algorithm for solving inequalities of the form (x - x 1)(x - x 2)…(x - x n) > 0 or (x - x 1)(x - x 2)…(x - x n)< 0, где x 1 , x 2 , … x n не равные друг другу числа.
The algorithm for solving inequalities using the interval method is displayed on the interactive whiteboard:
III. Learning new material. Solving fractional rational inequalities with multiple roots using the interval method.
Solving inequalities with multiple critical values of a variable is usually associated with the greatest difficulties. If previously it was possible to place signs on intervals simply by alternating them, now, when passing through a critical value, the sign of the entire expression may not change. We will get acquainted with the so-called “petal” method, which will help overcome the difficulties associated with arranging the signs of a function on intervals.
Consider an example: (x+3) 2 > 0/
The left side has a single critical point x = - 3. Let's mark it on the number line. This point has a multiplicity of 2, so we can consider that we have two merged critical points, between which there is also an interval with the beginning and end at the same point -3. We will mark such intervals with “petals”, as in Fig. 3. Thus, we have three intervals: two numerical intervals (-∞; -3); (-3; +∞) and the “petal” between them. All that remains is to place the signs. To do this, we calculate the sign on the interval containing zero, and arrange the signs on the rest, simply alternating them. The result of placing the signs is shown in Fig. 4
 |
 |
|
Rice. 3 |
Rice. 4 |
Answer: x € (-∞; -3) U (-3; +∞)
Let us now consider a more complex inequality (Fig. 5):
Let's introduce the function (Fig. 6):
Let's mark the critical points on the number line, taking into account their multiplicity - for each additional bracket with a given critical value we draw an additional “petal”. So, in Fig. 7, one “petal” will appear at the point x=3, since (x-3)?=(x-3)(x-3).
Since (x - 6) 3 = (x - 6) (x - 6) (x - 6), the point x = 6 has two "petals". The first multiplier is taken into account by point 6 on the axis, and two additional multipliers are taken into account by adding two “petals”. Next, we determine the sign on one of the intervals and arrange the signs on the rest, alternating minuses and pluses.
All spaces marked with a “+” and dark dots provide the answer.
X € [-4;-1) U (3) U (6;+∞).
IV. Consolidating new material
1. Let's solve the inequality:
Let us factor the left side of the inequality:
First, we plot the critical points of the denominator on the coordinate axis, we get (Fig. 10)
Adding numerator points, we get (Fig. 11)
And now, we determine the signs at intervals and in “petals” (Fig. 12)
Rice. 12
Answer: x € (-1; 0) U (0; 1) U (2)
2. Choose numerical intervals that are solutions to inequalities using the interval method, taking into account the multiplicity of the roots of the polynomial (Fig. 13).
V. Summary of the lesson
During the conversation with the class, we draw conclusions:
1) It becomes possible to place signs at intervals simply by alternating them.
3) With this solution, single roots are never lost.
In this lesson we will continue solving rational inequalities using the interval method for more complex inequalities. Let us consider the solution of fractional linear and fractional quadratic inequalities and related problems.
Now let's return to the inequality
Let's look at some related tasks.
Find the smallest solution to the inequality.
Find the number of natural solutions to the inequality
Find the length of the intervals that make up the set of solutions to the inequality.
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1. Mordkovich A.G. and others. Algebra 9th grade: Problem book for students of general education institutions / A. G. Mordkovich, T. N. Mishustina, etc. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 28(b,c); 29(b,c); 35(a,b); 37(b,c); 38(a).