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“Rational equations with polynomials” is one of the most common topics in the Unified State Exam test tasks 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 and bringing the equation to a common denominator, which is why completing such tasks causes difficulties. Solving rational equations in preparation for the Unified State Exam on our website will help you quickly cope with problems of any complexity and pass the test with flying colors.
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To know the rules for calculating unknowns and easily obtain correct results, use our online service. The Shkolkovo portal is a one-of-a-kind platform where materials necessary for preparing for the Unified State 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 standard rational equations, the basis of which is constantly updated and expanded.
For more effective preparation for testing, we recommend following our special method and starting with repeating the rules and solving simple problems, gradually moving on to more complex ones. Thus, the graduate will be able to identify the most difficult topics for himself and focus on studying them.
Start preparing for the final test with Shkolkovo today, and the results will not be long in coming! Choose the easiest example from those given. If you master the expression quickly, move on to a more difficult task. This way you can improve your knowledge up to the point of solving USE tasks in mathematics at a specialized level.
Training 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!
We have already learned how to solve quadratic equations. Now let's extend the studied methods to rational equations.
What is a rational expression? We have already encountered this concept. Rational expressions are expressions made up of numbers, variables, their powers and symbols of mathematical operations.
Accordingly, rational equations are equations of the form: , where 
- rational expressions.
Previously, we considered only those rational equations that can be reduced to linear ones. Now let's look at those rational equations that can be reduced to quadratic equations.
Example 1
Solve the equation: .
Solution:
A fraction is equal to 0 if and only if its numerator is equal to 0 and its denominator is not equal to 0.
We get the following system:
The first equation of the system is a quadratic equation. Before solving it, let's divide all its coefficients by 3. We get:
We get two roots: ; .
Since 2 never equals 0, two conditions must be met: 
. Since none of the roots of the equation obtained above coincides with 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 terms to the left side so that the right side ends up with 0.
2. Transform and simplify the left side, bring all fractions to a common denominator.
3. Equate the resulting fraction to 0 using the following algorithm: 
.
4. Write down those roots that were obtained in the first equation and satisfy the second inequality in the answer.
Let's look at another example.
Example 2
Solve the equation: .
Solution
At the very beginning, we move all the terms to the left so that 0 remains on the right. We get:
Now let's 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.
Coefficients of this equation: . We calculate the discriminant:
We get two roots: ; .
Now let's 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 find 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 reduce to quadratic equations.
In the next lesson we will look at rational equations as models of real situations, and also look at motion problems.
Bibliography
- Bashmakov M.I. Algebra, 8th grade. - M.: Education, 2004.
- Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra, 8. 5th ed. - M.: Education, 2010.
- Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra, 8th grade. Textbook for general education institutions. - M.: Education, 2006.
- Festival of pedagogical ideas "Open Lesson" ().
- School.xvatit.com ().
- Rudocs.exdat.com ().
Homework
Equations with fractions themselves are not difficult and are very interesting. Let's look at the types of fractional equations and how 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 the 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 fractions. Multiply each term in the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 are cancelled, 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 one. There are two possible solutions here.
- Option 1: We get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90×10 => 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. Divide 10 by 10, multiply by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.
We often encounter fractional equations in which the x's are on opposite sides of the equal sign. In such situations, it is necessary to move all the fractions with X's to one side, and the numbers to the other.
- 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. Specifying these conditions is a mandatory and integral part of a correct decision. By not adding 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 the unknown, a, b, c are ordinary numbers. Please note that x may not be any number. For example, x cannot equal zero, since it cannot be divided by 0. This is precisely the additional condition that we must specify. This is called the range of permissible values, abbreviated as VA.
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 fractions. 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, for example, 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 move -3 to the right side, changing the “-” sign to “+” and we get that x ≠ 3. The ODZ is indicated.
We solve the equation, multiply everything by x-3: 15 + 18×(x – 3) = 21×(x – 3) => 15 + 18x – 54 = 21x – 63.
Move the X's to the right, numbers to the left: 24 = 3x => x = 8.
Let's continue talking about solving equations. In this article we will go into detail about rational equations and principles of solving rational equations with one variable. First, let's figure out what type of equations are called rational, give a definition of whole rational and fractional rational equations, and give examples. Next, we will obtain algorithms for solving rational equations, and, of course, we will consider solutions to typical examples with all the necessary explanations.
Page navigation.
Based on the stated definitions, we give several examples of rational equations. For example, x=1, 2·x−12·x 2 ·y·z 3 =0, , are all rational equations.
From the examples shown, it is clear that rational equations, as well as equations of other types, can be with one variable, or with two, three, etc. variables. In the following paragraphs we will talk about solving rational equations with one variable. Solving equations in two variables and their large number deserve special attention.
In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional ones. Let us give the corresponding definitions.
Definition.
The rational equation is called whole, if both its left and right sides are integer rational expressions.
Definition.
If at least one of the parts of a rational equation is a fractional expression, then such an equation is called fractionally rational(or fractional rational).
It is clear that whole equations do not contain division by a variable; on the contrary, fractional rational equations necessarily contain division by a variable (or a variable in the denominator). So 3 x+2=0 and (x+y)·(3·x 2 −1)+x=−y+0.5– these are whole rational equations, both of their parts are whole expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.
Concluding this point, let us pay attention to the fact that the linear equations and quadratic equations known to this point are entire rational equations.
Solving whole equations
One of the main approaches to solving entire equations is to reduce them to equivalent ones algebraic equations. This can always be done by performing the following equivalent transformations of the equation:
- first, the expression from the right side of the original integer equation is transferred to the left side with the opposite sign to obtain zero on the right side;
- after this, on the left side of the equation the resulting standard form.
The result is an algebraic equation that is equivalent to the original integer equation. Thus, in the simplest cases, solving entire equations is reduced to solving linear or quadratic equations, and in the general case, to solving an algebraic equation of degree n. For clarity, let's look at the solution to the example.
Example.
Find the roots of the whole equation 3·(x+1)·(x−3)=x·(2·x−1)−3.
Solution.
Let us reduce the solution of this entire equation to the solution of an equivalent algebraic equation. To do this, firstly, we transfer the expression from the right side to the left, as a result we arrive at the equation 3·(x+1)·(x−3)−x·(2·x−1)+3=0. And, secondly, we transform the expression formed on the left side into a standard form polynomial by completing the necessary: 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. Thus, solving the original integer equation is reduced to solving the quadratic equation x 2 −5·x−6=0.
We calculate its discriminant D=(−5) 2 −4·1·(−6)=25+24=49, it is positive, which means that the equation has two real roots, which we find using the formula for the roots of a quadratic equation:
To be completely sure, let's do it checking the found roots of the equation. First we check the root 6, substitute it instead of the variable x in the original integer equation: 3·(6+1)·(6−3)=6·(2·6−1)−3, which is the same, 63=63. This is a valid numerical equation, therefore x=6 is indeed the root of the equation. Now we check the root −1, we have 3·(−1+1)·(−1−3)=(−1)·(2·(−1)−1)−3, from where, 0=0 . When x=−1, the original equation also turns into a correct numerical equality, therefore, x=−1 is also a root of the equation.
Answer:
6 , −1 .
Here it should also be noted that the term “degree of the whole equation” is associated with the representation of an entire equation in the form of an algebraic equation. Let us give the corresponding definition:
Definition.
The power of the whole equation is called the degree of an equivalent algebraic equation.
According to this definition, the entire equation from the previous example has the second degree.
This could have been the end of solving entire rational equations, if not for one thing…. As is known, solving algebraic equations of degree above the second is associated with significant difficulties, and for equations of degree above the fourth there are no general root formulas at all. Therefore, to solve entire equations of the third, fourth and higher degrees, it is often necessary to resort to other solution methods.
In such cases, an approach to solving entire rational equations based on factorization method. In this case, the following algorithm is adhered to:
- first, they ensure that there is a zero on the right side of the equation; to do this, they transfer the expression from the right side of the whole equation to the left;
- then, the resulting expression on the left side is presented as a product of several factors, which allows us to move on to a set of several simpler equations.
The given algorithm for solving an entire equation through factorization requires a detailed explanation using an example.
Example.
Solve the whole equation (x 2 −1)·(x 2 −10·x+13)= 2 x (x 2 −10 x+13) .
Solution.
First, as usual, we transfer the expression from the right side to the left side of the equation, not forgetting to change the sign, we get (x 2 −1)·(x 2 −10·x+13)− 2 x (x 2 −10 x+13)=0 . Here it is quite obvious that it is not advisable to transform the left-hand side of the resulting equation into a polynomial of the standard form, since this will give an algebraic equation of the fourth degree of the form x 4 −12 x 3 +32 x 2 −16 x−13=0, the solution of which is difficult.
On the other hand, it is obvious that on the left side of the resulting equation we can x 2 −10 x+13 , thereby presenting it as a product. We have (x 2 −10 x+13) (x 2 −2 x−1)=0. The resulting equation is equivalent to the original whole equation, and it, in turn, can be replaced by a set of two quadratic equations x 2 −10·x+13=0 and x 2 −2·x−1=0. Finding their roots using known root formulas through a discriminant is not difficult; the roots are equal. They are the desired roots of the original equation.
Answer:
Also useful for solving entire rational equations method for introducing a new variable. In some cases, it allows you to move to equations whose degree is lower than the degree of the original whole equation.
Example.
Find the real roots of a rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).
Solution.
Reducing this whole rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.
Here it is easy to see that you can introduce a new variable y and replace the expression x 2 +3·x with it. This replacement leads us to the whole equation (y+1) 2 +10=−2·(y−4) , which, after moving the expression −2·(y−4) to the left side and subsequent transformation of the expression formed there, is reduced to a quadratic equation y 2 +4·y+3=0. The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be selected based on the theorem inverse to Vieta's theorem.
Now we move on to the second part of the method of introducing a new variable, that is, to performing a reverse replacement. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3, which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . Using the formula for the roots of a quadratic equation, we find the roots of the first equation. And the second quadratic equation has no real roots, since its discriminant is negative (D=3 2 −4·3=9−12=−3 ).
Answer:
In general, when we are dealing with entire equations of high degrees, we must always be prepared to search for a non-standard method or an artificial technique for solving them.
Solving fractional rational equations
First, it will be useful to understand how to solve fractional rational equations of the form , where p(x) and q(x) are integer rational expressions. And then we will show how to reduce the solution of other fractionally rational equations to the solution of equations of the indicated type.
One approach to solving the equation is based on the following statement: the numerical fraction u/v, where v is a non-zero number (otherwise we will encounter , which is undefined), is equal to zero if and only if its numerator is equal to zero, then is, if and only if u=0 . By virtue of this statement, solving the equation is reduced to fulfilling two conditions p(x)=0 and q(x)≠0.
This conclusion corresponds to the following algorithm for solving a fractional rational equation. To solve a fractional rational equation of the form , you need
- solve the whole rational equation p(x)=0 ;
- and check whether the condition q(x)≠0 is satisfied for each root found, while
- if true, then this root is the root of the original equation;
- if it is not satisfied, then this root is extraneous, that is, it is not the root of the original equation.
Let's look at an example of using the announced algorithm when solving a fractional rational equation.
Example.
Find the roots of the equation.
Solution.
This is a fractional rational equation, and of the form , where p(x)=3·x−2, q(x)=5·x 2 −2=0.
According to the algorithm for solving fractional rational equations of this type, we first need to solve the equation 3 x−2=0. This is a linear equation whose root is x=2/3.
It remains to check for this root, that is, check whether it satisfies the condition 5 x 2 −2≠0. We substitute the number 2/3 into the expression 5 x 2 −2 instead of x, and we get . The condition is met, so x=2/3 is the root of the original equation.
Answer:
2/3 .
You can approach solving a fractional rational equation from a slightly different position. This equation is equivalent to the integer equation p(x)=0 on the variable x of the original equation. That is, you can stick to this algorithm for solving a fractional rational equation :
- solve the equation p(x)=0 ;
- find the ODZ of variable x;
- take roots belonging to the region of acceptable values - they are the desired roots of the original fractional rational equation.
For example, let's solve a fractional rational equation using this algorithm.
Example.
Solve the equation.
Solution.
First, we solve the quadratic equation x 2 −2·x−11=0. Its roots can be calculated using the root formula for the even second coefficient, we have D 1 =(−1) 2 −1·(−11)=12, And .
Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3·x≠0, which is the same as x·(x+3)≠0, whence x≠0, x≠−3.
It remains to check whether the roots found in the first step are included in the ODZ. Obviously yes. Therefore, the original fractional rational equation has two roots.
Answer:
Note that this approach is more profitable than the first if the ODZ is easy to find, and is especially beneficial if the roots of the equation p(x) = 0 are irrational, for example, or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and −31/59. This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational effort, and it is easier to exclude extraneous roots using the ODZ.
In other cases, when solving the equation, especially when the roots of the equation p(x) = 0 are integers, it is more profitable to use the first of the given algorithms. That is, it is advisable to immediately find the roots of the entire equation p(x)=0, and then check whether the condition q(x)≠0 is satisfied for them, rather than finding the ODZ, and then solving the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to check than to find DZ.
Let us consider the solution of two examples to illustrate the specified nuances.
Example.
Find the roots of the equation.
Solution.
First, let's find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, composed using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this 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 . Three of these equations are linear and one is quadratic; we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.
With the roots found, it is quite easy to check whether the denominator of the fraction on the left side of the original equation vanishes, but determining the ODZ, on the contrary, is not so simple, since for this you will have to solve an algebraic equation of the fifth degree. Therefore, we will abandon finding the ODZ in favor of checking the roots. To do this, we substitute them one by one instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (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 .
Thus, 1/2, 6 and −2 are the desired roots of the original fractional rational equation, and 7 and −1 are extraneous roots.
Answer:
1/2 , 6 , −2 .
Example.
Find the roots of a fractional rational equation.
Solution.
First, let's find the roots of the equation (5 x 2 −7 x−1) (x−2)=0. This equation is equivalent to a set of two equations: square 5 x 2 −7 x−1=0 and linear x−2=0. Using the formula for the roots of a quadratic equation, we find two roots, and from the second equation we have x=2.
Checking whether the denominator goes to zero at the found values of x is quite unpleasant. And determining the range of permissible values of the variable x in the original equation is quite simple. Therefore, we will act through ODZ.
In our case, the ODZ of the variable x of the original fractional rational equation consists of all numbers except those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we draw a conclusion about the ODZ: it consists of all x such that .
It remains to check whether the found roots and x=2 belong to the range of acceptable values. The roots belong, therefore, they are roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.
Answer:
It will also be useful to separately dwell on the cases when in a fractional rational equation of the form there is a number in the numerator, that is, when p(x) is represented by some number. Wherein
- if this number is non-zero, then the equation has no roots, since a fraction is equal to zero if and only if its numerator is equal to zero;
- if this number is zero, then the root of the equation is any number from the ODZ.
Example.
Solution.
Since the numerator of the fraction on the left side of the equation contains a non-zero number, then for any x the value of this fraction cannot be equal to zero. Therefore, this equation has no roots.
Answer:
no roots.
Example.
Solve the equation.
Solution.
The numerator of the fraction on the left side of this fractional rational equation contains zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the ODZ of this variable.
It remains to determine this range of acceptable values. It includes all values of x for which x 4 +5 x 3 ≠0. The solutions to the equation 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 combination of two equations x 3 =0 and x +5=0, from where these roots are visible. Therefore, the desired range of acceptable values is any x except x=0 and x=−5.
Thus, a fractional rational equation has infinitely many solutions, which are any numbers except zero and minus five.
Answer:
Finally, it's time to talk about solving fractional rational equations of arbitrary form. 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. Looking ahead, let's say that their solution comes down to solving equations of the form already familiar to us.
It is known that transferring a term from one part of the equation to another with the opposite sign leads to an equivalent equation, therefore the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0.
We also know that any , identically equal to this expression, is possible. Thus, we can always transform the rational expression on the left side of the equation r(x)−s(x)=0 into an identically equal rational fraction of the form .
So we move from the original fractional rational equation r(x)=s(x) to the equation, and its solution, as we found out above, reduces to solving the equation p(x)=0.
But here it is necessary to take into account the fact that when replacing r(x)−s(x)=0 with , and then with p(x)=0, the range of permissible values of the variable x may expand.
Consequently, the original equation r(x)=s(x) and the equation p(x)=0 that we arrived at may turn out to be unequal, and by solving the equation p(x)=0, we can get roots that will be extraneous roots of the original equation r(x)=s(x) . You can identify and not include extraneous roots in the answer either by performing a check or by checking that they belong to the ODZ of the original equation.
Let's summarize this information in algorithm for solving fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , you need
- Get zero on the right by moving the expression from the right side with the opposite sign.
- Perform operations with fractions and polynomials on the left side of the equation, thereby transforming it into a rational fraction of the form.
- Solve the equation p(x)=0.
- Identify and eliminate extraneous roots, which is done by substituting them into the original equation or by checking their belonging to the ODZ of the original equation.
For greater clarity, we will show the entire chain of solving fractional rational equations:
.
Let's look at the solutions of several examples with a detailed explanation of the solution process in order to clarify the given block of information.
Example.
Solve a fractional rational equation.
Solution.
We will act in accordance with the solution algorithm just obtained. And first we move the terms from the right side of the equation to the left, as a result we move on to the equation.
In the second step, we need to convert the fractional rational expression on the left side of the resulting equation to the form of a fraction. To do this, we reduce rational fractions to a common denominator and simplify the resulting expression: . So we come to the equation.
In the next step, we need to solve the equation −2·x−1=0. We find x=−1/2.
It remains to check whether the found number −1/2 is not an extraneous root of the original equation. To do this, you can check or find the VA of the variable x of the original equation. Let's demonstrate both approaches.
Let's start with checking. We substitute the number −1/2 into the original equation instead of the variable x, and we get the same thing, −1=−1. The substitution gives the correct numerical equality, so x=−1/2 is the root of the original equation.
Now we will show how the last point of the algorithm is performed through ODZ. The range of permissible values of the original equation is the set of all numbers except −1 and 0 (at x=−1 and x=0 the denominators of the fractions vanish). The root x=−1/2 found in the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.
Answer:
−1/2 .
Let's look at another example.
Example.
Find the roots of the equation.
Solution.
We need to solve a fractional rational equation, let's go through all the steps of the algorithm.
First, we move the term from the right side to the left, we get .
Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0.
Its root is obvious - it is zero.
At the fourth step, it remains to find out whether the found root is extraneous to the original fractional rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it doesn't make sense because it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.
7, which leads to Eq. From this we can conclude that the expression in the denominator of the left side must be equal to that of the right side, that is, . Now we subtract from both sides of the triple: . By analogy, from where, and further.
The check shows that both roots found are roots of the original fractional rational equation.
Answer:
Bibliography.
- Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
- Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.