We have already learned how to solve quadratic equations. Let us now extend the studied methods to rational equations.

What is a rational expression? We have already encountered this concept. Rational expressions called expressions made up of numbers, variables, their degrees and signs of mathematical operations.

Accordingly, rational equations are equations of the form: , where - rational expressions.

Previously, we considered only those rational equations that reduce to linear ones. Now let's consider those rational equations that can be reduced to quadratic ones.

Example 1

Solve the equation: .

Decision:

A fraction is 0 if and only if its numerator is 0 and its denominator is not 0.

We get the following system:

The first equation of the system is a quadratic equation. Before solving it, we divide all its coefficients by 3. We get:

We get two roots: ; .

Since 2 is never equal to 0, two conditions must be met: . Since none of the roots of the equation obtained above matches the invalid values ​​of the variable that were obtained when solving the second inequality, they are both solutions to this equation.

Answer:.

So, let's formulate an algorithm for solving rational equations:

1. Move all the terms to the left side so that 0 is obtained on the right side.

2. Transform and simplify the left side, bring all fractions to a common denominator.

3. Equate the resulting fraction to 0, according to the following algorithm: .

4. Write down those roots that are obtained in the first equation and satisfy the second inequality in response.

Let's look at another example.

Example 2

Solve the equation: .

Decision

At the very beginning, we transfer all the terms to the left side so that 0 remains on the right. We get:

Now we bring the left side of the equation to a common denominator:

This equation is equivalent to the system:

The first equation of the system is a quadratic equation.

The coefficients of this equation: . We calculate the discriminant:

We get two roots: ; .

Now we solve the second inequality: the product of factors is not equal to 0 if and only if none of the factors is equal to 0.

Two conditions must be met: . We get that of the two roots of the first equation, only one is suitable - 3.

Answer:.

In this lesson, we remembered what a rational expression is, and also learned how to solve rational equations, which are reduced to quadratic equations.

In the next lesson, we will consider rational equations as models of real situations, and also consider motion problems.

Bibliography

1. Bashmakov M.I. Algebra, 8th grade. - M.: Enlightenment, 2004.
2. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. et al. Algebra, 8. 5th ed. - M.: Education, 2010.
3. Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for educational institutions. - M.: Education, 2006.

1. Festival of pedagogical ideas "Open Lesson" ().
2. School.xvatit.com().
3. Rudocs.exdat.com().

Homework

"Rational equations with polynomials" is one of the most frequently encountered topics in the USE tests in mathematics. For this reason, their repetition should be given special attention. Many students are faced with the problem of finding the discriminant, transferring indicators from the right side to the left side and bringing the equation to a common denominator, which makes it difficult to complete such tasks. Solving rational equations in preparation for the exam on our website will help you quickly cope with tasks of any complexity and pass the test perfectly.

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To know the rules for calculating unknowns and easily get the correct results, use our online service. The Shkolkovo portal is a one-of-a-kind platform where the materials necessary for preparing for the exam are collected. Our teachers systematized and presented in an understandable form all the mathematical rules. In addition, we invite schoolchildren to try their hand at solving typical rational equations, the base of which is constantly updated and supplemented.

For more effective preparation for testing, we recommend that you follow our special method and start by repeating the rules and solving simple problems, gradually moving on to more complex ones. Thus, the graduate will be able to highlight the most difficult topics for himself and focus on their study.

Start preparing for the final testing with Shkolkovo today, and the result will not keep you waiting! Choose the easiest example from those given. If you quickly mastered the expression, move on to a more difficult task. So you can improve your knowledge up to solving USE tasks in mathematics at the profile level.

Education is available not only to graduates from Moscow, but also to schoolchildren from other cities. Spend a couple of hours a day studying on our portal, for example, and very soon you will be able to cope with equations of any complexity!

So far, we have only solved integer equations with respect to the unknown, that is, equations in which the denominators (if any) did not contain the unknown.

Often you have to solve equations that contain the unknown in the denominators: such equations are called fractional.

To solve this equation, we multiply both sides of it by that is, by a polynomial containing the unknown. Will the new equation be equivalent to the given one? To answer the question, let's solve this equation.

Multiplying both sides of it by , we get:

Solving this equation of the first degree, we find:

So, equation (2) has a single root

Substituting it into equation (1), we get:

Hence, is also the root of equation (1).

Equation (1) has no other roots. In our example, this can be seen, for example, from the fact that in equation (1)

How the unknown divisor must be equal to the dividend 1 divided by the quotient 2, i.e.

So, equations (1) and (2) have a single root. Hence, they are equivalent.

2. We now solve the following equation:

The simplest common denominator: ; multiply all the terms of the equation by it:

After reduction we get:

Let's expand the brackets:

Bringing like terms, we have:

Solving this equation, we find:

Substituting into equation (1), we get:

On the left side, we received expressions that do not make sense.

Hence, the root of equation (1) is not. This implies that equations (1) and are not equivalent.

In this case, we say that equation (1) has acquired an extraneous root.

Let us compare the solution of equation (1) with the solution of the equations we considered earlier (see § 51). In solving this equation, we had to perform two such operations that had not been seen before: first, we multiplied both sides of the equation by an expression containing the unknown (common denominator), and, second, we reduced algebraic fractions by factors containing the unknown .

Comparing Equation (1) with Equation (2), we see that not all x values ​​valid for Equation (2) are valid for Equation (1).

It is the numbers 1 and 3 that are not admissible values ​​of the unknown for equation (1), and as a result of the transformation they became admissible for equation (2). One of these numbers turned out to be a solution to equation (2), but, of course, it cannot be a solution to equation (1). Equation (1) has no solutions.

This example shows that when multiplying both parts of the equation by a factor containing the unknown, and when reducing algebraic fractions, an equation can be obtained that is not equivalent to the given one, namely: extraneous roots can appear.

Hence we draw the following conclusion. When solving an equation containing an unknown in the denominator, the resulting roots must be checked by substitution into the original equation. Extraneous roots must be discarded.

Equations with fractions themselves are not difficult and very interesting. Consider the types of fractional equations and ways to solve them.

How to solve equations with fractions - x in the numerator

If a fractional equation is given, where the unknown is in the numerator, the solution does not require additional conditions and is solved without unnecessary hassle. The general form of such an equation is x/a + b = c, where x is an unknown, a, b and c are ordinary numbers.

Find x: x/5 + 10 = 70.

In order to solve the equation, you need to get rid of the fractions. Multiply each term of the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 is reduced, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 - 50 = 300.

Find x: x/5 + x/10 = 90.

This example is a slightly more complicated version of the first. There are two solutions here.

• Option 1: Get rid of fractions by multiplying all terms of the equation by the larger denominator, i.e. by 10: 10x/5 + 10x/10 = 90x10 => 2x + x = 900 => 3x = 900 => x=300.
• Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. 10 divided by 10, multiplied by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.

Often there are fractional equations in which x's are on opposite sides of the equal sign. In such a situation, it is necessary to transfer all fractions with x in one direction, and the numbers in another.

• Find x: 3x/5 = 130 - 2x/5.
• Move 2x/5 to the right with the opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
• We reduce 5x/5 and get: x = 130.

How to solve an equation with fractions - x in the denominator

This type of fractional equations requires writing additional conditions. The indication of these conditions is a mandatory and integral part of the right decision. By not attributing them, you run the risk, since the answer (even if it is correct) may simply not be counted.

The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is an unknown, a, b, c are ordinary numbers. Note that x may not be any number. For example, x cannot be zero, since you cannot divide by 0. This is precisely the additional condition that we must specify. This is called the range of acceptable values, abbreviated - ODZ.

Find x: 15/x + 18 = 21.

We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation according to the standard scheme, getting rid of the fractions. We multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.

Often there are equations where the denominator contains not only x, but also some other operation with it, such as addition or subtraction.

Find x: 15/(x-3) + 18 = 21.

We already know that the denominator cannot be equal to zero, which means x-3 ≠ 0. We transfer -3 to the right side, while changing the “-” sign to “+” and we get that x ≠ 3. ODZ is indicated.

Solve the equation, multiply everything by x-3: 15 + 18x(x - 3) = 21x(x - 3) => 15 + 18x - 54 = 21x - 63.

Move the x's to the right, the numbers to the left: 24 = 3x => x = 8.

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 their 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 look at 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 standard form polynomial.

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 on the left side into a polynomial of the standard form and perform the 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, there are no general formulas for the roots at all. 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

First, let's solve the 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 is simpler than the first one in cases where the area of ​​​​admissible values ​​of the variable x is easily found, 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) .

To do this, we have to reduce 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, with the exception of − 1 and 0 (for 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

It will be easiest 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

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