Interval method is a universal way to solve almost any inequalities that appear in a school algebra course. It is based on the following properties of functions:
1. A continuous function g(x) can change sign only at the point at which it is equal to 0. Graphically, this means that the graph of a continuous function can move from one half-plane to another only if it intersects the x-axis (we remember that the ordinate of any point lying on the OX axis (abscissa axis) is equal to zero, that is, the value of the function at this point is equal to 0):
We see that the function y=g(x) shown on the graph intersects the OX axis at points x= -8, x=-2, x=4, x=8. These points are called zeros of the function. And at the same points the function g(x) changes sign.
2. The function can also change the sign at the zeros of the denominator - simplest example well known function:
We see that the function changes sign at the root of the denominator, at point , but does not vanish at any point. Thus, if a function contains a fraction, it can change sign at the roots of the denominator.
2. However, the function does not always change sign at the root of the numerator or at the root of the denominator. For example, the function y=x 2 does not change sign at the point x=0:
Because the equation x 2 =0 has two equal roots x=0, at the point x=0 the function seems to turn to 0 twice. Such a root is called a root of the second multiplicity.
Function 
changes the sign at zero of the numerator, , but does not change the sign at zero of the denominator: , since the root is the root of the second multiplicity, that is, of even multiplicity:
Important! In roots of even multiplicity the function does not change sign.
Note! Any nonlinear Inequalities in school algebra courses are usually solved using the method of intervals.
I offer you a detailed one, following which you can avoid mistakes when solving nonlinear inequalities.
1. First you need to bring the inequality to the form
P(x)V0,
where V is the inequality sign:<,>,≤ or ≥. To do this you need:
a) move all terms to the left side of the inequality,
b) find the roots of the resulting expression,
c) factor the left side of the inequality
d) write identical factors as powers.
Attention! The last step must be done in order not to make a mistake with the multiplicity of the roots - if the result is a multiplier to an even power, then the corresponding root has an even multiplicity.
2. Plot the found roots on the number axis.
3. If the inequality is strict, then the circles indicating the roots on the number axis are left “empty”; if the inequality is not strict, then the circles are filled in.
4. We select roots of even multiplicity - in them P(x) the sign does not change.
5. Determine the sign P(x) on the rightmost gap. To do this, take an arbitrary value x 0, which is greater than the larger root and substitute it into P(x).
If P(x 0)>0 (or ≥0), then in the rightmost space we put a “+” sign.
If P(x 0)<0 (или ≤0), то в самом правом промежутке ставим знак "-".
When passing through the point indicating a root of even multiplicity, the sign DOES NOT CHANGE.
7. Once again we look at the sign of the original inequality, and select the intervals of the sign we need.
8. Attention! If our inequality is NOT STRICT, then we check the condition of equality to zero separately.
9. Write down the answer.
If the original the inequality contains an unknown in the denominator, then we also move all terms to the left, and reduce the left side of the inequality to the form
(where V is the inequality sign:< или >)
A strict inequality of this type is equivalent to the inequality
NOT Strict inequality of the form
equivalent system:
In practice, if the function has the form , then we proceed as follows:
- Find the roots of the numerator and denominator.
- We apply them to the axle. Leave all circles empty. Then, if the inequality is not strict, then we paint over the roots of the numerator, and always leave the roots of the denominator empty.
- Next we follow the general algorithm:
- We select roots of even multiplicity (if the numerator and denominator contain the same roots, then we count how many times the same roots occur). In roots of even multiplicity, the sign does not change.
- We find out the sign on the rightmost gap.
- We are putting up signs.
- In the case of a non-strict inequality, we check the condition of equality and the condition of equality to zero separately.
- We select the necessary gaps and free-standing roots.
- We write down the answer.
To better understand algorithm for solving inequalities using the interval method, watch the VIDEO TUTORIAL, which explains the example in detail solving inequalities using the interval method.
Lesson topic "Solving systems of rational inequalities"
Class 10
Lesson type: search
Goal: finding ways to solve inequalities with modulus, applying the interval method in a new situation.
Lesson objectives:
Test your skills in solving rational inequalities and their systems; - show students the possibility of using the interval method when solving inequalities with modulus;
Teach to think logically;
Develop the skill of self-assessment of your work;
Learn to express your thoughts
Learn to defend your point of view with reason;
To form a positive motive for learning in students;
Develop student independence.
During the classes
I. Organizing time(1 min)
Hello, today we will continue to study the topic “System of rational inequalities”, we will apply our knowledge and skills in a new situation.
Write down the date and topic of the lesson "Solving systems of rational inequalities." Today I invite you on a journey along the roads of mathematics, where tests await you, a test of strength. On your desks are road maps with tasks, a self-assessment travel sheet, which you will hand over to me (the dispatcher) at the end of the trip.
The motto of the trip will be the aphorism “He who walks can master the road, but he who thinks in mathematics”. Take your knowledge with you. Engage your thought process and hit the road. On the road we will be accompanied by a road radio.A fragment of music plays (1 min). Then a sharp sound of a signal.
II. Knowledge testing stage. Work in groups."Baggage Inspection"
Here comes the first Baggage Screening test, testing your knowledge on the topic
Now you will be divided into groups of 3 or 4 people. Everyone has a piece of paper with a task on their desk. Distribute these tasks among each other, solve them, and write down the ready-made answers on a common sheet. A group of 3 people chooses any 3 tasks. Anyone who completes all tasks will report this to the teacher. I or my assistants will check the answers, and if at least one answer is incorrect, the group will be returned a sheet for rechecking. (children do not see the answers, they are only told which task has the wrong answer).The winner is the group that is the first to complete all tasks without errors. Forward to victory.
The music is very quiet.
If two or three groups finish their work at the same time, one of the children from the other group will help the teacher check. Answers on the teacher's sheet (4 copies).
Work stops when the winning group appears.
Don't forget to complete the self-assessment worksheet. And we move on.
Task sheet for “Baggage Inspection”
1) 3)
2) 4)
III. The stage of updating knowledge and discovering new knowledge. "Eureka"
The inspection showed that you have a wealth of knowledge.
But on the road all sorts of situations happen, sometimes ingenuity is required, and we’ll check if you forgot to take it with you.
You have learned to solve systems of rational inequalities using the interval method. Today we will look at which problems it is advisable to use this method. But first, let's remember what a module is.
1. Continue the sentences “The modulus of a number is equal to the number itself if...”(orally)
“The modulus of a number is equal to the opposite number if...”
2. Let A(X) be a polynomial in x
Continue recording:
Answer:
Write down the opposite expression of A(x)
A(x) = 5 - 4x; A(x) = 6x 2 - 4x + 2
A(x)= -A(x)=
The student writes on the board, the guys write in their notebooks.
3. Now let’s try to find a way to solve the quadratic inequality with modulus
What are your suggestions for solving this inequality?
Listen to the guys' suggestions.
If there are no proposals, then ask the question: “Can this inequality be solved using systems of inequalities?”
The student comes out and decides.
IV. The stage of primary consolidation of new knowledge, drawing up a solution algorithm. Baggage replenishment.
(Work in groups of 4 people).
Now I suggest you replenish your luggage. You will work in groups.Each group is given 2 task cards.
On the first card you need to write down systems for solving the inequalities presented on the board and develop an algorithm for solving such inequalities; there is no need to solve them.
The first card is different for the groups, the second is the same
What happened?
Under each equation on the board you need to write a set of systems.
4 students come out and write systems. At this time, we discuss the algorithm with the class.
V. Stage of consolidation of knowledge."Way home".
The luggage is replenished, now it’s time to head back. Now solve any of the proposed inequalities with modulus yourself in accordance with the compiled algorithm.
The road radio will again be with you on the road.
Play quiet background music. The teacher checks the design and provides advice if necessary.
Tasks on the board.
The work has been completed. Check the answers (they are on back side boards), fill out the self-assessment travel sheet.
Setting homework.
Write it down homework(copy in your notebook the inequalities that you did not do or did with errors, additionally No. 84 (a) on page 373 of the textbook if desired)
VI. Relaxation stage.
How was this trip useful for you?
What have you learned?
Summarize. Count how many points each of you earned.(the guys name the final score).Hand over the self-assessment sheets to the dispatcher, that is, to me.
I want to end the lesson with a parable.
“A sage walked, and three people met him, carrying carts with stones for construction under the hot sun. The sage stopped and asked each one a question. He asked the first one: “What have you been doing all day?”, and he answered with a grin that he had been carrying the damned stones all day. The sage asked the second: “What did you do all day?”, and he answered: “I did my job conscientiously,” and the third smiled, his face lit up with joy and pleasure: “And I took part in the construction of the Temple!”
The lesson is over.
Self-assessment sheet
Last name, first name, class | Number of points |
|
Working in a group to solve inequalities or systems of inequalities. | 2 points if done correctly without outside help; 1 point if done correctly with outside help; 0 points if you didn’t complete the task 1 extra point for group victory |
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.
Yandex.RTB R-A-339285-1
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.
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 integer 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 let’s mark everything on the diagram 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. Required interval equals (− 0 , 5 , 6) , therefore, this range of values will be the solution we need.
Answer: (− 0 , 5 , 6) .
There are more complex cases, when the left turns out to be a polynomial of third or higher 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:
- 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 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 give 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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