"solution of fractional rational equations". Rational Equations

The lowest common denominator is used to simplify given equation. This method is used when you cannot write the given equation with one rational expression on each side of the equation (and use the cross multiplication method). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, cross multiplication is better).

Find the least common denominator of fractions (or least common multiple). NOZ is smallest number, which is evenly divisible by each denominator.

• Sometimes NOZ is an obvious number. For example, if the equation is given: x/3 + 1/2 = (3x + 1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 will be 6.
• If the NOD is not obvious, write down the multiples of the largest denominator and find among them one that is a multiple of the other denominators as well. You can often find the NOD by simply multiplying two denominators together. For example, if the equation x/8 + 2/6 = (x - 3)/9 is given, then NOZ = 8*9 = 72.
• If one or more denominators contain a variable, then the process is somewhat more complicated (but not impossible). In this case, the NOZ is an expression (containing a variable) that is divisible by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divisible by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).

Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOZ by the corresponding denominator of each fraction. Since you're multiplying both the numerator and denominator by the same number, you're effectively multiplying a fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).

• So in our example, multiply x/3 by 2/2 to get 2x/6, and multiply 1/2 by 3/3 to get 3/6 (3x + 1/6 does not need to be multiplied because it the denominator is 6).
• Proceed similarly when the variable is in the denominator. In our second example NOZ = 3x(x-1), so 5/(x-1) times (3x)/(3x) is 5(3x)/(3x)(x-1); 1/x times 3(x-1)/3(x-1) to get 3(x-1)/3x(x-1); 2/(3x) multiply by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).

Find x. Now that you've reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by a common denominator. Then solve the resulting equation, that is, find "x". To do this, isolate the variable on one side of the equation.

• In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
• In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by NOZ, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.

Simply put, these are equations in which there is at least one with a variable in the denominator.

For example:

\(\frac(9x^2-1)(3x)\) \(=0\)
\(\frac(1)(2x)+\frac(x)(x+1)=\frac(1)(2)\)
\(\frac(6)(x+1)=\frac(x^2-5x)(x+1)\)

Example not fractional rational equations:

\(\frac(9x^2-1)(3)\) \(=0\)
\(\frac(x)(2)\) \(+8x^2=6\)

How are fractional rational equations solved?

The main thing to remember about fractional rational equations is that you need to write in them. And after finding the roots, be sure to check them for admissibility. Otherwise, extraneous roots may appear, and the whole solution will be considered incorrect.

Algorithm for solving a fractional rational equation:

Write out and "solve" the ODZ.

Multiply each term in the equation by a common denominator and reduce the resulting fractions. The denominators will disappear.

Write the equation without opening brackets.

Solve the resulting equation.

Check the found roots with ODZ.

Write down in response the roots that passed the test in step 7.

Do not memorize the algorithm, 3-5 solved equations - and it will be remembered by itself.

Example . Solve fractional rational equation \(\frac(x)(x-2) - \frac(7)(x+2)=\frac(8)(x^2-4)\)

Decision:

Answer: \(3\).

Example . Find the roots of the fractional rational equation \(=0\)

Decision:

\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)(x^2+7x+10)\)\(=0\) ODZ: \(x+2≠0⇔x≠-2\) \(x+5≠0 ⇔x≠-5\) \(x^2+7x+10≠0\) \(D=49-4 \cdot 10=9\) \(x_1≠\frac(-7+3)(2)=-2\) \(x_2≠\frac(-7-3)(2)=-5\)		We write down and "solve" ODZ. Expand \(x^2+7x+10\) into the formula: \(ax^2+bx+c=a(x-x_1)(x-x_2)\). Fortunately \(x_1\) and \(x_2\) we have already found.
\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)((x+2)(x+5))\)\(=0\)		Obviously, the common denominator of fractions: \((x+2)(x+5)\). We multiply the whole equation by it.
\(\frac(x(x+2)(x+5))(x+2) + \frac((x+1)(x+2)(x+5))(x+5)-\) \(-\frac((7-x)(x+2)(x+5))((x+2)(x+5))\)\(=0\)		We reduce fractions
\(x(x+5)+(x+1)(x+2)-7+x=0\)		Opening the brackets
\(x^2+5x+x^2+3x+2-7+x=0\)		We give like terms
\(2x^2+9x-5=0\)		Finding the roots of the equation
\(x_1=-5;\) \(x_2=\frac(1)(2).\)		One of the roots does not fit under the ODZ, so in response we write down only the second root.

Answer: \(\frac(1)(2)\).

Lesson Objectives:

Tutorial:

• formation of the concept of fractional rational equations;
• to consider various ways of solving fractional rational equations;
• consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
• to teach the solution of fractional rational equations according to the algorithm;
• checking the level of assimilation of the topic by conducting test work.

Developing:

• development of the ability to correctly operate with the acquired knowledge, to think logically;
• development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization;
• development of initiative, the ability to make decisions, not to stop there;
• development critical thinking;
• development of research skills.

Nurturing:

• upbringing cognitive interest to the subject;
• education of independence in solving educational problems;
• education of will and perseverance to achieve the final results.

Lesson type: lesson - explanation of new material.

During the classes

1. Organizational moment.

Hello guys! The equations are written on the blackboard, look at them carefully. Can you solve all of these equations? Which ones are not and why?

Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study today in the lesson? Formulate the topic of the lesson. So, we open notebooks and write down the topic of the lesson “Solution of fractional rational equations”.

2. Actualization of knowledge. Frontal survey, oral work with the class.

And now we will repeat the main theoretical material that we need to study new topic. Please answer the following questions:

1. What is an equation? ( Equality with a variable or variables.)
2. What is equation #1 called? ( Linear.) Method for solving linear equations. ( Move everything with the unknown to the left side of the equation, all numbers to the right. Bring like terms. Find the unknown multiplier).
3. What is Equation 3 called? ( Square.) Methods for solving quadratic equations. ( Selection of the full square, by formulas, using the Vieta theorem and its consequences.)
4. What is a proportion? ( Equality of two relations.) The main property of proportion. ( If the proportion is true, then the product of its extreme terms is equal to the product of the middle terms.)
5. What properties are used to solve equations? ( 1. If in the equation we transfer the term from one part to another, changing its sign, then we get an equation equivalent to the given one. 2. If both parts of the equation are multiplied or divided by the same non-zero number, then an equation will be obtained that is equivalent to the given.)
6. When is a fraction equal to zero? ( The fraction is zero when the numerator zero, and the denominator is not equal to zero.)

3. Explanation of new material.

Solve equation No. 2 in notebooks and on the board.

Answer: 10.

What fractional rational equation can you try to solve using the basic property of proportion? (No. 5).

(x-2)(x-4) = (x+2)(x+3)

x 2 -4x-2x + 8 \u003d x 2 + 3x + 2x + 6

x 2 -6x-x 2 -5x \u003d 6-8

Solve equation No. 4 in notebooks and on the board.

Answer: 1,5.

What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).

x 2 -7x+12 = 0

D=1>0, x 1 =3, x 2 =4.

Answer: 3;4.

Now try to solve equation #7 in one of the ways.

(x 2 -2x-5)x(x-5)=x(x-5)(x+5)
(x 2 -2x-5)x(x-5)-x(x-5)(x+5)=0			x 2 -2x-5=x+5
x(x-5)(x 2 -2x-5-(x+5))=0			x 2 -2x-5-x-5=0
x(x-5)(x 2 -3x-10)=0
x=0 x-5=0 x 2 -3x-10=0
x 1 \u003d 0 x 2 \u003d 5 D \u003d 49
x 3 \u003d 5 x 4 \u003d -2			x 3 \u003d 5 x 4 \u003d -2
Answer: 0;5;-2.			Answer: 5;-2.

Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?

Until now, students have not met the concept of an extraneous root, it is really very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.

• How do equations No. 2 and 4 differ from equations No. 5,6,7? ( In equations No. 2 and 4 in the denominator of the number, No. 5-7 - expressions with a variable.)
• What is the root of the equation? ( The value of the variable at which the equation becomes a true equality.)
• How to find out if a number is the root of an equation? ( Make a check.)

When doing a test, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that eliminates this error? Yes, this method is based on the condition that the fraction is equal to zero.

x 2 -3x-10=0, D=49, x 1 =5, x 2 = -2.

If x=5, then x(x-5)=0, so 5 is an extraneous root.

If x=-2, then x(x-5)≠0.

Answer: -2.

Let's try to formulate an algorithm for solving fractional rational equations in this way. Children themselves formulate the algorithm.

Algorithm for solving fractional rational equations:

1. Move everything to the left.
2. Bring fractions to a common denominator.
3. Make up a system: a fraction is zero when the numerator is zero and the denominator is not zero.
4. Solve the equation.
5. Check inequality to exclude extraneous roots.
6. Write down the answer.

Discussion: how to formalize the solution if the basic property of proportion is used and the multiplication of both sides of the equation by a common denominator. (Supplement the solution: exclude from its roots those that turn the common denominator to zero).

4. Primary comprehension of new material.

Work in pairs. Students choose how to solve the equation on their own, depending on the type of equation. Tasks from the textbook "Algebra 8", Yu.N. Makarychev, 2007: No. 600 (b, c, i); No. 601(a, e, g). The teacher controls the performance of the task, answers the questions that have arisen, and provides assistance to poorly performing students. Self-test: Answers are written on the board.

b) 2 is an extraneous root. Answer:3.

c) 2 is an extraneous root. Answer: 1.5.

a) Answer: -12.5.

g) Answer: 1; 1.5.

5. Statement of homework.

1. Read item 25 from the textbook, analyze examples 1-3.
2. Learn the algorithm for solving fractional rational equations.
3. Solve in notebooks No. 600 (a, d, e); No. 601 (g, h).
4. Try to solve #696(a) (optional).

6. Fulfillment of the control task on the studied topic.

The work is done on sheets.

Job example:

A) Which of the equations are fractional rational?

B) A fraction is zero when the numerator is ______________________ and the denominator is _______________________.

Q) Is the number -3 the root of Equation #6?

D) Solve equation No. 7.

Task evaluation criteria:

• "5" is given if the student completed more than 90% of the task correctly.
• "4" - 75% -89%
• "3" - 50% -74%
• "2" is given to a student who completed less than 50% of the task.
• Grade 2 is not put in the journal, 3 is optional.

7. Reflection.

On the leaflets with independent work, put:

• 1 - if the lesson was interesting and understandable to you;
• 2 - interesting, but not clear;
• 3 - not interesting, but understandable;
• 4 - not interesting, not clear.

8. Summing up the lesson.

So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations different ways, tested their knowledge with the help of training independent work. You will learn the results of independent work in the next lesson, at home you will have the opportunity to consolidate the knowledge gained.

What method of solving fractional rational equations, in your opinion, is easier, more accessible, more rational? Regardless of the method of solving fractional rational equations, what should not be forgotten? What is the "cunning" of fractional rational equations?

Thank you all, the lesson is over.

Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.

Yandex.RTB R-A-339285-1

Rational Equation: Definition and Examples

Acquaintance with rational expressions begins in the 8th grade of the school. At this time, in algebra lessons, students are increasingly beginning to meet tasks with equations that contain rational expressions in your notes. Let's refresh our memory of what it is.

Definition 1

rational equation is an equation in which both sides contain rational expressions.

In various manuals, you can find another wording.

Definition 2

rational equation- this is an equation, the record of the left side of which contains a rational expression, and the right one contains zero.

The definitions that we have given for rational equations are equivalent, since they mean the same thing. The correctness of our words is confirmed by the fact that for any rational expressions P and Q equations P=Q and P − Q = 0 will be equivalent expressions.

Now let's turn to examples.

Example 1

Rational equations:

x = 1 , 2 x − 12 x 2 y z 3 = 0 , x x 2 + 3 x - 1 = 2 + 2 7 x - a (x + 2) , 1 2 + 3 4 - 12 x - 1 = 3 .

Rational equations, just like equations of other types, can contain any number of variables from 1 to several. To begin with, we will consider simple examples, in which the equations will contain only one variable. And then we begin to gradually complicate the task.

Rational equations are divided into two large groups: integer and fractional. Let's see which equations will apply to each of the groups.

Definition 3

A rational equation will be an integer if the record of its left and right parts contains entire rational expressions.

Definition 4

A rational equation will be fractional if one or both of its parts contain a fraction.

Fractionally rational equations necessarily contain division by a variable, or the variable is present in the denominator. There is no such division in writing integer equations.

Example 2

3 x + 2 = 0 and (x + y) (3 x 2 − 1) + x = − y + 0 , 5 are entire rational equations. Here both parts of the equation are represented by integer expressions.

1 x - 1 = x 3 and x: (5 x 3 + y 2) = 3: (x − 1) : 5 are fractionally rational equations.

Entire rational equations include linear and quadratic equations.

Solving integer equations

The solution of such equations usually reduces to their transformation into equivalent algebraic equations. This can be achieved by carrying out equivalent transformations of the equations in accordance with the following algorithm:

• first we get zero on the right side of the equation, for this it is necessary to transfer the expression that is on the right side of the equation to its left side and change the sign;
• then we transform the expression on the left side of the equation into a polynomial standard view.

We have to get an algebraic equation. This equation will be equivalent with respect to the original equation. Easy cases allow us to solve the problem by reducing the whole equation to a linear or quadratic one. In the general case, we solve an algebraic equation of degree n.

Example 3

It is necessary to find the roots of the whole equation 3 (x + 1) (x − 3) = x (2 x − 1) − 3.

Decision

Let us transform the original expression in order to obtain an algebraic equation equivalent to it. To do this, we will transfer the expression contained in the right side of the equation to the left side and change the sign to the opposite. As a result, we get: 3 (x + 1) (x − 3) − x (2 x − 1) + 3 = 0.

Now we will transform the expression that is on the left side into a polynomial of the standard form and perform necessary actions with this polynomial:

3 (x + 1) (x - 3) - x (2 x - 1) + 3 = (3 x + 3) (x - 3) - 2 x 2 + x + 3 = = 3 x 2 - 9 x + 3 x - 9 - 2 x 2 + x + 3 = x 2 - 5 x - 6

We managed to reduce the solution of the original equation to the solution of a quadratic equation of the form x 2 − 5 x − 6 = 0. The discriminant of this equation is positive: D = (− 5) 2 − 4 1 (− 6) = 25 + 24 = 49 . This means that there will be two real roots. Let's find them using the formula of the roots of the quadratic equation:

x \u003d - - 5 ± 49 2 1,

x 1 \u003d 5 + 7 2 or x 2 \u003d 5 - 7 2,

x 1 = 6 or x 2 = - 1

Let's check the correctness of the roots of the equation that we found in the course of the solution. For this number, which we received, we substitute into the original equation: 3 (6 + 1) (6 − 3) = 6 (2 6 − 1) − 3 and 3 (− 1 + 1) (− 1 − 3) = (− 1) (2 (− 1) − 1) − 3. In the first case 63 = 63 , in the second 0 = 0 . Roots x=6 and x = − 1 are indeed the roots of the equation given in the example condition.

Answer: 6 , − 1 .

Let's look at what "power of the whole equation" means. We will often come across this term in those cases when we need to represent an entire equation in the form of an algebraic one. Let's define the concept.

Definition 5

Degree of an integer equation is the degree of an algebraic equation equivalent to the original whole equation.

If you look at the equations from the example above, you can establish: the degree of this whole equation is the second.

If our course was limited to solving equations of the second degree, then the consideration of the topic could be completed here. But everything is not so simple. Solving equations of the third degree is fraught with difficulties. And for equations above the fourth degree, it does not exist at all general formulas roots. In this regard, the solution of entire equations of the third, fourth and other degrees requires us to use a number of other techniques and methods.

The most commonly used approach to solving entire rational equations is based on the factorization method. The algorithm of actions in this case is as follows:

• we transfer the expression from the right side to the left side so that zero remains on the right side of the record;
• we represent the expression on the left side as a product of factors, and then we move on to a set of several simpler equations.

Example 4

Find the solution to the equation (x 2 − 1) (x 2 − 10 x + 13) = 2 x (x 2 − 10 x + 13) .

Decision

We transfer the expression from the right side of the record to the left side with the opposite sign: (x 2 − 1) (x 2 − 10 x + 13) − 2 x (x 2 − 10 x + 13) = 0. Converting the left side to a polynomial of the standard form is impractical due to the fact that this will give us an algebraic equation of the fourth degree: x 4 − 12 x 3 + 32 x 2 − 16 x − 13 = 0. The ease of transformation does not justify all the difficulties with solving such an equation.

It is much easier to go the other way: we take out the common factor x 2 − 10 x + 13 . Thus we arrive at an equation of the form (x 2 − 10 x + 13) (x 2 − 2 x − 1) = 0. Now we replace the resulting equation with a set of two quadratic equations x 2 − 10 x + 13 = 0 and x 2 − 2 x − 1 = 0 and find their roots through the discriminant: 5 + 2 3 , 5 - 2 3 , 1 + 2 , 1 - 2 .

Answer: 5 + 2 3 , 5 - 2 3 , 1 + 2 , 1 - 2 .

Similarly, we can use the method of introducing a new variable. This method allows us to pass to equivalent equations with powers lower than those in the original whole equation.

Example 5

Does the equation have roots? (x 2 + 3 x + 1) 2 + 10 = − 2 (x 2 + 3 x − 4)?

Decision

If we now try to reduce a whole rational equation to an algebraic one, we will get an equation of degree 4, which has no rational roots. Therefore, it will be easier for us to go the other way: introduce a new variable y, which will replace the expression in the equation x 2 + 3 x.

Now we will work with the whole equation (y + 1) 2 + 10 = − 2 (y − 4). We transfer the right side of the equation to the left side with the opposite sign and carry out the necessary transformations. We get: y 2 + 4 y + 3 = 0. Let's find the roots of the quadratic equation: y = − 1 and y = − 3.

Now let's do the reverse substitution. We get two equations x 2 + 3 x = − 1 and x 2 + 3 x = - 3 . Let's rewrite them as x 2 + 3 x + 1 = 0 and x 2 + 3 x + 3 = 0. We use the formula of the roots of the quadratic equation in order to find the roots of the first equation obtained: - 3 ± 5 2 . The discriminant of the second equation is negative. This means that the second equation has no real roots.

Answer:- 3 ± 5 2

Integer equations of high degrees come across in problems quite often. There is no need to be afraid of them. You need to be ready to apply a non-standard method of solving them, including a number of artificial transformations.

Solution of fractionally rational equations

We begin our consideration of this subtopic with an algorithm for solving fractionally rational equations of the form p (x) q (x) = 0 , where p(x) and q(x) are integer rational expressions. The solution of other fractionally rational equations can always be reduced to the solution of equations of the indicated form.

The most commonly used method for solving equations p (x) q (x) = 0 is based on the following statement: numerical fraction u v, where v is a number that is different from zero, equal to zero only in cases where the numerator of the fraction is equal to zero. Following the logic of the above statement, we can assert that the solution of the equation p (x) q (x) = 0 can be reduced to the fulfillment of two conditions: p(x)=0 and q(x) ≠ 0. On this, an algorithm for solving fractional rational equations of the form p (x) q (x) = 0 is built:

• we find the solution of the whole rational equation p(x)=0;
• we check whether the condition is satisfied for the roots found during the solution q(x) ≠ 0.

If this condition is met, then the found root. If not, then the root is not a solution to the problem.

Example 6

Find the roots of the equation 3 · x - 2 5 · x 2 - 2 = 0 .

Decision

We are dealing with a fractional rational equation of the form p (x) q (x) = 0 , in which p (x) = 3 · x − 2 , q (x) = 5 · x 2 − 2 = 0 . Let's start solving the linear equation 3 x - 2 = 0. The root of this equation will be x = 2 3.

Let's check the found root, whether it satisfies the condition 5 x 2 - 2 ≠ 0. To do this, substitute a numeric value into the expression. We get: 5 2 3 2 - 2 \u003d 5 4 9 - 2 \u003d 20 9 - 2 \u003d 2 9 ≠ 0.

The condition is met. It means that x = 2 3 is the root of the original equation.

Answer: 2 3 .

There is another option for solving fractional rational equations p (x) q (x) = 0 . Recall that this equation is equivalent to the whole equation p(x)=0 on the range of admissible values ​​of the variable x of the original equation. This allows us to use the following algorithm in solving the equations p(x) q(x) = 0:

• solve the equation p(x)=0;
• find the range of acceptable values ​​for the variable x ;
• we take the roots that lie in the region of admissible values ​​of the variable x as the desired roots of the original fractional rational equation.

Example 7

Solve the equation x 2 - 2 x - 11 x 2 + 3 x = 0 .

Decision

To start, let's decide quadratic equation x 2 − 2 x − 11 = 0. To calculate its roots, we use the root formula for an even second coefficient. We get D 1 = (− 1) 2 − 1 (− 11) = 12, and x = 1 ± 2 3 .

Now we can find the ODV of x for the original equation. These are all numbers for which x 2 + 3 x ≠ 0. It's the same as x (x + 3) ≠ 0, whence x ≠ 0 , x ≠ − 3 .

Now let's check whether the roots x = 1 ± 2 3 obtained at the first stage of the solution are within the range of acceptable values ​​of the variable x . We see what comes in. This means that the original fractional rational equation has two roots x = 1 ± 2 3 .

Answer: x = 1 ± 2 3

The second solution method described easier than the first in cases where it is easy to find the area of ​​​​admissible values ​​of the variable x, and the roots of the equation p(x)=0 irrational. For example, 7 ± 4 26 9 . Roots can be rational, but with a large numerator or denominator. For example, 127 1101 and − 31 59 . This saves time for checking the condition. q(x) ≠ 0: it is much easier to exclude roots that do not fit, according to the ODZ.

When the roots of the equation p(x)=0 are integers, it is more expedient to use the first of the described algorithms for solving equations of the form p (x) q (x) = 0 . Finding the roots of an entire equation faster p(x)=0, and then check whether the condition is met for them q(x) ≠ 0, and not find the ODZ, and then solve the equation p(x)=0 on this ODZ. This is due to the fact that in such cases it is usually easier to make a check than to find the ODZ.

Example 8

Find the roots of the equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) x 5 - 15 x 4 + 57 x 3 - 13 x 2 + 26 x + 112 = 0 .

Decision

We start by considering the whole equation (2 x - 1) (x - 6) (x 2 - 5 x + 14) (x + 1) = 0 and finding its roots. To do this, we apply the method of solving equations through factorization. It turns out that the original equation is equivalent to a set of four equations 2 x - 1 = 0, x - 6 = 0, x 2 - 5 x + 14 = 0, x + 1 = 0, of which three are linear and one is square. We find the roots: from the first equation x = 1 2, from the second x=6, from the third - x \u003d 7, x \u003d - 2, from the fourth - x = − 1.

Let's check the obtained roots. It is difficult for us to determine the ODZ in this case, since for this we will have to solve an algebraic equation of the fifth degree. It will be easier to check the condition according to which the denominator of the fraction, which is on the left side of the equation, should not vanish.

In turn, substitute the roots in place of the variable x in the expression x 5 − 15 x 4 + 57 x 3 − 13 x 2 + 26 x + 112 and calculate its value:

1 2 5 - 15 1 2 4 + 57 1 2 3 - 13 1 2 2 + 26 1 2 + 112 = = 1 32 - 15 16 + 57 8 - 13 4 + 13 + 112 = 122 + 1 32 ≠0;

6 5 − 15 6 4 + 57 6 3 − 13 6 2 + 26 6 + 112 = 448 ≠ 0 ;

7 5 − 15 7 4 + 57 7 3 − 13 7 2 + 26 7 + 112 = 0 ;

(− 2) 5 − 15 (− 2) 4 + 57 (− 2) 3 − 13 (− 2) 2 + 26 (− 2) + 112 = − 720 ≠ 0 ;

(− 1) 5 − 15 (− 1) 4 + 57 (− 1) 3 − 13 (− 1) 2 + 26 (− 1) + 112 = 0 .

The verification carried out allows us to establish that the roots of the original fractional rational equation are 1 2 , 6 and − 2 .

Answer: 1 2 , 6 , - 2

Example 9

Find the roots of the fractional rational equation 5 x 2 - 7 x - 1 x - 2 x 2 + 5 x - 14 = 0 .

Decision

Let's start with the equation (5 x 2 - 7 x - 1) (x - 2) = 0. Let's find its roots. It is easier for us to represent this equation as a combination of quadratic and linear equations 5 x 2 - 7 x - 1 = 0 and x − 2 = 0.

We use the formula of the roots of a quadratic equation to find the roots. We get two roots x = 7 ± 69 10 from the first equation, and from the second x=2.

Substituting the value of the roots into the original equation to check the conditions will be quite difficult for us. It will be easier to determine the LPV of the variable x . In this case, the DPV of the variable x is all numbers, except for those for which the condition is satisfied x 2 + 5 x − 14 = 0. We get: x ∈ - ∞ , - 7 ∪ - 7 , 2 ∪ 2 , + ∞ .

Now let's check if the roots we found belong to the range of acceptable values ​​for the x variable.

The roots x = 7 ± 69 10 - belong, therefore, they are the roots of the original equation, and x=2- does not belong, therefore, it is an extraneous root.

Answer: x = 7 ± 69 10 .

Let us examine separately the cases when the numerator of a fractional rational equation of the form p (x) q (x) = 0 contains a number. In such cases, if the numerator contains a number other than zero, then the equation will not have roots. If this number is equal to zero, then the root of the equation will be any number from the ODZ.

Example 10

Solve the fractional rational equation - 3 , 2 x 3 + 27 = 0 .

Decision

This equation will not have roots, since the numerator of the fraction from the left side of the equation contains a non-zero number. This means that for any values ​​of x the value of the fraction given in the condition of the problem will not be equal to zero.

Answer: no roots.

Example 11

Solve the equation 0 x 4 + 5 x 3 = 0.

Decision

Since the numerator of the fraction is zero, the solution to the equation will be any value of x from the ODZ variable x.

Now let's define the ODZ. It will include all x values ​​for which x 4 + 5 x 3 ≠ 0. Equation solutions x 4 + 5 x 3 = 0 are 0 and − 5 , since this equation is equivalent to the equation x 3 (x + 5) = 0, and it, in turn, is equivalent to the set of two equations x 3 = 0 and x + 5 = 0 where these roots are visible. We come to the conclusion that the desired range of acceptable values ​​are any x , except x=0 and x = -5.

It turns out that the fractional rational equation 0 x 4 + 5 x 3 = 0 has an infinite number of solutions, which are any numbers except zero and - 5.

Answer: - ∞ , - 5 ∪ (- 5 , 0 ∪ 0 , + ∞

Now let's talk about fractional rational equations of an arbitrary form and methods for solving them. They can be written as r(x) = s(x), where r(x) and s(x) are rational expressions, and at least one of them is fractional. The solution of such equations is reduced to the solution of equations of the form p (x) q (x) = 0 .

We already know that we can get an equivalent equation by transferring the expression from the right side of the equation to the left side with the opposite sign. This means that the equation r(x) = s(x) is equivalent to the equation r (x) − s (x) = 0. We have also already discussed how to convert a rational expression into a rational fraction. Thanks to this, we can easily transform the equation r (x) − s (x) = 0 into its identical rational fraction of the form p (x) q (x) .

So we move from the original fractional rational equation r(x) = s(x) to an equation of the form p (x) q (x) = 0 , which we have already learned how to solve.

It should be noted that when making transitions from r (x) − s (x) = 0 to p (x) q (x) = 0 and then to p(x)=0 we may not take into account the expansion of the range of valid values ​​of the variable x .

It is quite realistic that the original equation r(x) = s(x) and equation p(x)=0 as a result of the transformations, they will cease to be equivalent. Then the solution of the equation p(x)=0 can give us roots that will be foreign to r(x) = s(x). In this regard, in each case it is necessary to carry out a check by any of the methods described above.

To make it easier for you to study the topic, we have generalized all the information into an algorithm for solving a fractional rational equation of the form r(x) = s(x):

• we transfer the expression from the right side with the opposite sign and get zero on the right;
• we transform the original expression into a rational fraction p (x) q (x) by sequentially performing actions with fractions and polynomials;
• solve the equation p(x)=0;
• we reveal extraneous roots by checking their belonging to the ODZ or by substituting into the original equation.

Visually, the chain of actions will look like this:

r (x) = s (x) → r (x) - s (x) = 0 → p (x) q (x) = 0 → p (x) = 0 → dropout r o n d e r o o n s

Example 12

Solve the fractional rational equation x x + 1 = 1 x + 1 .

Decision

Let's move on to the equation x x + 1 - 1 x + 1 = 0 . Let's transform the fractional rational expression on the left side of the equation to the form p (x) q (x) .

For this we have to bring rational fractions to a common denominator and simplify the expression:

x x + 1 - 1 x - 1 = x x - 1 (x + 1) - 1 x (x + 1) x (x + 1) = = x 2 - x - 1 - x 2 - x x (x + 1) = - 2 x - 1 x (x + 1)

In order to find the roots of the equation - 2 x - 1 x (x + 1) = 0, we need to solve the equation − 2 x − 1 = 0. We get one root x = - 1 2.

It remains for us to perform the check by any of the methods. Let's consider them both.

Substitute the resulting value into the original equation. We get - 1 2 - 1 2 + 1 = 1 - 1 2 + 1 . We have come to the correct numerical equality − 1 = − 1 . It means that x = − 1 2 is the root of the original equation.

Now we will check through the ODZ. Let's determine the area of ​​acceptable values ​​for the variable x . This will be the entire set of numbers, except for − 1 and 0 (when x = − 1 and x = 0, the denominators of fractions vanish). The root we got x = − 1 2 belongs to the ODZ. This means that it is the root of the original equation.

Answer: − 1 2 .

Example 13

Find the roots of the equation x 1 x + 3 - 1 x = - 2 3 x .

Decision

We are dealing with a fractional rational equation. Therefore, we will act according to the algorithm.

Let's move the expression from the right side to the left side with the opposite sign: x 1 x + 3 - 1 x + 2 3 x = 0

Let's carry out the necessary transformations: x 1 x + 3 - 1 x + 2 3 x = x 3 + 2 x 3 = 3 x 3 = x.

We come to the equation x=0. The root of this equation is zero.

Let's check if this root is a foreign one for the original equation. Substitute the value in the original equation: 0 1 0 + 3 - 1 0 = - 2 3 0 . As you can see, the resulting equation does not make sense. This means that 0 is an extraneous root, and the original fractional rational equation has no roots.

Answer: no roots.

If we have not included other equivalent transformations in the algorithm, this does not mean at all that they cannot be used. The algorithm is universal, but it is designed to help, not limit.

Example 14

Solve the equation 7 + 1 3 + 1 2 + 1 5 - x 2 = 7 7 24

Decision

The easiest way is to solve the given fractional rational equation according to the algorithm. But there is another way. Let's consider it.

Subtract from the right and left parts 7, we get: 1 3 + 1 2 + 1 5 - x 2 \u003d 7 24.

From this we can conclude that the expression in the denominator of the left side should be equal to the number reciprocal of the number from the right side, that is, 3 + 1 2 + 1 5 - x 2 = 24 7 .

Subtract from both parts 3: 1 2 + 1 5 - x 2 = 3 7 . By analogy 2 + 1 5 - x 2 \u003d 7 3, from where 1 5 - x 2 \u003d 1 3, and further 5 - x 2 \u003d 3, x 2 \u003d 2, x \u003d ± 2

Let's check in order to establish whether the found roots are the roots of the original equation.

Answer: x = ± 2

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

We introduced the equation above in § 7. First, we recall what a rational expression is. This is - algebraic expression, composed of numbers and the variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.

If r(x) is a rational expression, then the equation r(x) = 0 is called a rational equation.

However, in practice it is more convenient to use somewhat more broad interpretation term "rational equation": this is an equation of the form h(x) = q(x), where h(x) and q(x) are rational expressions.

Until now, we could not solve any rational equation, but only one that, as a result of various transformations and reasoning, was reduced to linear equation. Now our possibilities are much greater: we will be able to solve a rational equation, which reduces not only to linear
mu, but also to the quadratic equation.

Recall how we solved rational equations earlier and try to formulate a solution algorithm.

Example 1 solve the equation

Decision. We rewrite the equation in the form

In this case, as usual, we use the fact that the equalities A \u003d B and A - B \u003d 0 express the same relationship between A and B. This allowed us to transfer the term to the left side of the equation with the opposite sign.

Let's perform transformations of the left side of the equation. We have

Recall the equality conditions fractions zero: if, and only if, two relations are simultaneously satisfied:

1) the numerator of the fraction is zero (a = 0); 2) the denominator of the fraction is different from zero).
Equating to zero the numerator of the fraction on the left side of equation (1), we obtain

It remains to check the fulfillment of the second condition mentioned above. The ratio means for equation (1) that . The values ​​x 1 = 2 and x 2 = 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.

1) Let's transform the equation to the form

2) Let's perform the transformations of the left side of this equation:

(simultaneously changed the signs in the numerator and
fractions).
Thus, given equation takes the form

3) Solve the equation x 2 - 6x + 8 = 0. Find

4) For the found values, check the condition . The number 4 satisfies this condition, but the number 2 does not. So 4 is the root of the given equation, and 2 is an extraneous root.
Answer: 4.

2. Solution of rational equations by introducing a new variable

The method of introducing a new variable is familiar to you, we have used it more than once. Let us show by examples how it is used in solving rational equations.

Example 3 Solve the equation x 4 + x 2 - 20 = 0.

Decision. We introduce a new variable y \u003d x 2. Since x 4 \u003d (x 2) 2 \u003d y 2, then the given equation can be rewritten in the form

y 2 + y - 20 = 0.

This is a quadratic equation, the roots of which we will find using the known formulas; we get y 1 = 4, y 2 = - 5.
But y \u003d x 2, which means that the problem has been reduced to solving two equations:
x2=4; x 2 \u003d -5.

From the first equation we find the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 + c \u003d 0 is called a biquadratic equation (“bi” - two, i.e., as it were, a “twice square” equation). The equation just solved was exactly biquadratic. Any biquadratic equation is solved in the same way as the equation from example 3: a new variable y \u003d x 2 is introduced, the resulting quadratic equation is solved with respect to the variable y, and then returned to the variable x.

Example 4 solve the equation

Decision. Note that the same expression x 2 + 3x occurs twice here. Hence, it makes sense to introduce a new variable y = x 2 + Zx. This will allow us to rewrite the equation in a simpler and more pleasant form (which, in fact, is the purpose of introducing a new variable- and recording is easier
, and the structure of the equation becomes clearer):

And now we will use the algorithm for solving a rational equation.

1) Let's move all the terms of the equation into one part:

= 0
2) Let's transform the left side of the equation

So, we have transformed the given equation into the form

3) From the equation - 7y 2 + 29y -4 = 0 we find (we have already solved quite a lot of quadratic equations, so it’s probably not worth always giving detailed calculations in the textbook).

4) Let's check the found roots using the condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved:
Since y \u003d x 2 + Zx, and y, as we have established, takes two values: 4 and, - we still have to solve two equations: x 2 + Zx \u003d 4; x 2 + Zx \u003d. The roots of the first equation are the numbers 1 and - 4, the roots of the second equation are the numbers

In the examples considered, the method of introducing a new variable was, as mathematicians like to say, adequate to the situation, that is, it corresponded well to it. Why? Yes, because the same expression was clearly encountered in the equation several times and it was reasonable to designate this expression with a new letter. But this is not always the case, sometimes a new variable "appears" only in the process of transformations. This is exactly what will happen in the next example.

Example 5 solve the equation
x(x-1)(x-2)(x-3) = 24.
Decision. We have
x (x - 3) \u003d x 2 - 3x;
(x - 1) (x - 2) \u003d x 2 -3x + 2.

So the given equation can be rewritten as

(x 2 - 3x)(x 2 + 3x + 2) = 24

Now a new variable has "appeared": y = x 2 - Zx.

With its help, the equation can be rewritten in the form y (y + 2) \u003d 24 and then y 2 + 2y - 24 \u003d 0. The roots of this equation are the numbers 4 and -6.

Returning to the original variable x, we obtain two equations x 2 - Zx \u003d 4 and x 2 - Zx \u003d - 6. From the first equation we find x 1 \u003d 4, x 2 \u003d - 1; the second equation has no roots.

Answer: 4, - 1.

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