The root of which equation is a fraction. The simplest rational equations
Solving equations with fractions let's look at examples. The examples are simple and illustrative. With their help, you can understand in the most understandable way,.
For example, you need to solve a simple equation x/b + c = d.
An equation of this type is called linear, because the denominator contains only numbers.
The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side is reduced.
For example, how to solve fractional equation:
x/5+4=9
We multiply both parts by 5. We get:
x+20=45
x=45-20=25
Another example where the unknown is in the denominator:
Equations of this type are called fractional rational or simply fractional.
We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic equation, which is solved in the usual way. You should only take into account the following points:
- the value of a variable that turns the denominator to 0 cannot be a root;
- you cannot divide or multiply the equation by the expression =0.
Here comes into force such a concept as the area of permissible values (ODZ) - these are the values \u200b\u200bof the roots of the equation for which the equation makes sense.
Thus, solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our DHS are excluded from the answer.
For example, you need to solve a fractional equation:
Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x - any value other than zero.
We get rid of the denominator by multiplying all terms of the equation by x
And solve the usual equation
5x - 2x = 1
3x=1
x = 1/3
Answer: x = 1/3
Let's solve the equation more complicated:
ODZ is also present here: x -2.
Solving this equation, we will not transfer everything in one direction and bring fractions to a common denominator. We immediately multiply both sides of the equation by an expression that will reduce all the denominators at once.
To reduce the denominators, you need to multiply the left side by x + 2, and the right side by 2. So, both sides of the equation must be multiplied by 2 (x + 2):
This is the most common multiplication of fractions, which we have already discussed above.
We write the same equation, but in a slightly different way.
The left side is reduced by (x + 2), and the right side by 2. After the reduction, we get the usual linear equation:
x \u003d 4 - 2 \u003d 2, which corresponds to our ODZ
Answer: x = 2.
Solving equations with fractions not as difficult as it might seem. In this article, we have shown this with examples. If you are having any difficulty with how to solve equations with fractions, then unsubscribe in the comments.
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Introduction to irrational equations
Guys, we learned how to solve quadratic equations. But mathematics is not limited to them. Today we will learn how to solve rational equations. concept rational equations very similar to the concept rational numbers. Only in addition to numbers, now we have introduced some variable $x$. And thus we get an expression in which there are operations of addition, subtraction, multiplication, division and raising to an integer power.Let $r(x)$ be rational expression
. Such an expression can be a simple polynomial in the variable $x$ or a ratio of polynomials (the operation of division is introduced, as for rational numbers).
The equation $r(x)=0$ is called rational equation.
Any equation of the form $p(x)=q(x)$, where $p(x)$ and $q(x)$ are rational expressions, will also be rational equation.
Consider examples of solving rational equations.
Example 1Solve the equation: $\frac(5x-3)(x-3)=\frac(2x-3)(x)$.
Solution.
Let's move all expressions to the left side: $\frac(5x-3)(x-3)-\frac(2x-3)(x)=0$.
If ordinary numbers were represented on the left side of the equation, then we would bring two fractions to a common denominator.
Let's do this: $\frac((5x-3)*x)((x-3)*x)-\frac((2x-3)*(x-3))((x-3)*x )=\frac(5x^2-3x-(2x^2-6x-3x+9))((x-3)*x)=\frac(3x^2+6x-9)((x-3) *x)=\frac(3(x^2+2x-3))((x-3)*x)$.
We got the equation: $\frac(3(x^2+2x-3))((x-3)*x)=0$.
A fraction is equal to zero if and only if the numerator of the fraction zero, and the denominator is different from zero. Then separately equate the numerator to zero and find the roots of the numerator.
$3(x^2+2x-3)=0$ or $x^2+2x-3=0$.
$x_(1,2)=\frac(-2±\sqrt(4-4*(-3)))(2)=\frac(-2±4)(2)=1;-3$.
Now let's check the denominator of the fraction: $(x-3)*x≠0$.
The product of two numbers is equal to zero when at least one of these numbers is equal to zero. Then: $x≠0$ or $x-3≠0$.
$x≠0$ or $x≠3$.
The roots obtained in the numerator and denominator do not match. So in response we write down both roots of the numerator.
Answer: $x=1$ or $x=-3$.
If suddenly, one of the roots of the numerator coincided with the root of the denominator, then it should be excluded. Such roots are called extraneous!
Algorithm for solving rational equations:
1. All expressions contained in the equation should be transferred to left side from the equal sign.2. Convert this part of the equation to algebraic fraction: $\frac(p(x))(q(x))=0$.
3. Equate the resulting numerator to zero, that is, solve the equation $p(x)=0$.
4. Equate the denominator to zero and solve the resulting equation. If the roots of the denominator coincided with the roots of the numerator, then they should be excluded from the answer.
Example 2
Solve the equation: $\frac(3x)(x-1)+\frac(4)(x+1)=\frac(6)(x^2-1)$.
Solution.
We will solve according to the points of the algorithm.
1. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=0$.
2. $\frac(3x)(x-1)+\frac(4)(x+1)-\frac(6)(x^2-1)=\frac(3x)(x-1)+\ frac(4)(x+1)-\frac(6)((x-1)(x+1))= \frac(3x(x+1)+4(x-1)-6)((x -1)(x+1))=$ $=\frac(3x^2+3x+4x-4-6)((x-1)(x+1))=\frac(3x^2+7x- 10)((x-1)(x+1))$.
$\frac(3x^2+7x-10)((x-1)(x+1))=0$.
3. Equate the numerator to zero: $3x^2+7x-10=0$.
$x_(1,2)=\frac(-7±\sqrt(49-4*3*(-10)))(6)=\frac(-7±13)(6)=-3\frac( 1)(3);1$.
4. Equate the denominator to zero:
$(x-1)(x+1)=0$.
$x=1$ and $x=-1$.
One of the roots $x=1$ coincided with the root of the numerator, then we do not write it down in response.
Answer: $x=-1$.
It is convenient to solve rational equations using the change of variables method. Let's demonstrate it.
Example 3
Solve the equation: $x^4+12x^2-64=0$.
Solution.
We introduce a replacement: $t=x^2$.
Then our equation will take the form:
$t^2+12t-64=0$ is an ordinary quadratic equation.
$t_(1,2)=\frac(-12±\sqrt(12^2-4*(-64)))(2)=\frac(-12±20)(2)=-16; 4$.
Let's introduce an inverse replacement: $x^2=4$ or $x^2=-16$.
The roots of the first equation are a pair of numbers $x=±2$. The second one has no roots.
Answer: $x=±2$.
Example 4
Solve the equation: $x^2+x+1=\frac(15)(x^2+x+3)$.
Solution.
Let's introduce a new variable: $t=x^2+x+1$.
Then the equation will take the form: $t=\frac(15)(t+2)$.
Next, we will act according to the algorithm.
1. $t-\frac(15)(t+2)=0$.
2. $\frac(t^2+2t-15)(t+2)=0$.
3. $t^2+2t-15=0$.
$t_(1,2)=\frac(-2±\sqrt(4-4*(-15)))(2)=\frac(-2±\sqrt(64))(2)=\frac( -2±8)(2)=-5; 3$.
4. $t≠-2$ - the roots do not match.
We introduce a reverse substitution.
$x^2+x+1=-5$.
$x^2+x+1=3$.
Let's solve each equation separately:
$x^2+x+6=0$.
$x_(1,2)=\frac(-1±\sqrt(1-4*(-6)))(2)=\frac(-1±\sqrt(-23))(2)$ - no roots.
And the second equation: $x^2+x-2=0$.
Rooted given equation there will be numbers $x=-2$ and $x=1$.
Answer: $x=-2$ and $x=1$.
Example 5
Solve the equation: $x^2+\frac(1)(x^2) +x+\frac(1)(x)=4$.
Solution.
We introduce a replacement: $t=x+\frac(1)(x)$.
Then:
$t^2=x^2+2+\frac(1)(x^2)$ or $x^2+\frac(1)(x^2)=t^2-2$.
We got the equation: $t^2-2+t=4$.
$t^2+t-6=0$.
The roots of this equation are the pair:
$t=-3$ and $t=2$.
Let's introduce the reverse substitution:
$x+\frac(1)(x)=-3$.
$x+\frac(1)(x)=2$.
We will decide separately.
$x+\frac(1)(x)+3=0$.
$\frac(x^2+3x+1)(x)=0$.
$x_(1,2)=\frac(-3±\sqrt(9-4))(2)=\frac(-3±\sqrt(5))(2)$.
Let's solve the second equation:
$x+\frac(1)(x)-2=0$.
$\frac(x^2-2x+1)(x)=0$.
$\frac((x-1)^2)(x)=0$.
The root of this equation is the number $x=1$.
Answer: $x=\frac(-3±\sqrt(5))(2)$, $x=1$.
Tasks for independent solution
Solve Equations:1. $\frac(3x+2)(x)=\frac(2x+3)(x+2)$.
2. $\frac(5x)(x+2)-\frac(20)(x^2+2x)=\frac(4)(x)$.
3. $x^4-7x^2-18=0$.
4. $2x^2+x+2=\frac(8)(2x^2+x+4)$.
5. $(x+2)(x+3)(x+4)(x+5)=3$.
We introduced the equation above in § 7. First, we recall what a rational expression is. This - 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
Solution. 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 relations 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.
Solution. 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
Solution. 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 record 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.
Solution. 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.
Let's get acquainted with rational and fractional rational equations, give their definition, give examples, and also analyze the most common types of problems.
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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 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 entire 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.
Solution
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 quadratic equation kind 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 algebraic equation, which is 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.
Find the solution to the equation (x 2 − 1) (x 2 − 10 x + 13) = 2 x (x 2 − 10 x + 13) .
Solution
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)?
Solution
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 the equations p (x) q (x) = 0 is based on the following statement: the 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 .
Solution
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.
Solve the equation x 2 - 2 x - 11 x 2 + 3 x = 0 .
Solution
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 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 .
Solution
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 .
Solution
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 .
Solution
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.
Solution
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) , sequentially performing operations 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 .
Solution
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, 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 .
Solution
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
Solution
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
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In this article I will show you algorithms for solving seven types of rational equations, which are reduced to square ones by means of a change of variables. In most cases, the transformations that lead to the replacement are very nontrivial, and it is quite difficult to guess about them on your own.
For each type of equation, I will explain how to make a change of variable in it, and then in the corresponding video tutorial I will show a detailed solution.
You have the opportunity to continue solving the equations yourself, and then check your solution with the video tutorial.
So, let's begin.
1 . (x-1)(x-7)(x-4)(x+2)=40
Note that the product of four brackets is on the left side of the equation, and the number is on the right side.
1. Let's group the brackets by two so that the sum of the free terms is the same.
2. Multiply them.
3. Let us introduce a change of variable.
In our equation, we group the first bracket with the third, and the second with the fourth, since (-1) + (-4) \u003d (-7) + 2:
At this point, the variable change becomes obvious:
We get the equation 
Answer: 
2 .
An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of a number by. And it is solved in a completely different way:
1. We group the brackets by two so that the product of the free terms is the same.
2. We multiply each pair of brackets.
3. From each factor, we take x out of the bracket.
4. Divide both sides of the equation by .
5. We introduce a change of variable.
In this equation, we group the first bracket with the fourth, and the second with the third, since:
Note that in each bracket the coefficient at and the free term are the same. Let's take out the multiplier from each bracket:
Since x=0 is not the root of the original equation, we divide both sides of the equation by . We get:
We get the equation: 
Answer: 
3
. 
Note that the denominators of both fractions contain square trinomials, whose leading coefficient and free term are the same. We take out, as in the equation of the second type, x out of the bracket. We get:
Divide the numerator and denominator of each fraction by x:
Now we can introduce a change of variable:
We get the equation for the variable t:
4 .
Note that the coefficients of the equation are symmetric with respect to the central one. Such an equation is called returnable .
To solve it
1. Divide both sides of the equation by (We can do this since x=0 is not the root of the equation.) We get:
2. Group the terms in this way:
3. In each group, we take out the common factor:
4. Let's introduce a replacement:
5. Let's express the expression in terms of t:
From here 
We get the equation for t:
Answer:
5. Homogeneous equations.
Equations that have the structure of a homogeneous one can be encountered when solving exponential, logarithmic and trigonometric equations, so it needs to be recognized.
Homogeneous equations have the following structure:
In this equality, A, B and C are numbers, and the same expressions are indicated by a square and a circle. That is, on the left side of the homogeneous equation is the sum of monomials that have the same degree (in this case, the degree of monomials is 2), and there is no free term.
To solve the homogeneous equation, we divide both sides by
Attention! When dividing the right and left sides of the equation by an expression containing an unknown, you can lose the roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both parts of the equation are the roots of the original equation.
Let's go the first way. We get the equation:
Now we introduce a variable substitution:
Simplify the expression and get a biquadratic equation for t:
Answer: or
7
. 
This equation has the following structure:
To solve it, you need to select the full square on the left side of the equation.
To select a full square, you need to add or subtract the double product. Then we get the square of the sum or the difference. This is critical to a successful variable substitution.
Let's start by finding the double product. It will be the key to replace the variable. In our equation, the double product is
Now let's figure out what is more convenient for us to have - the square of the sum or difference. Consider, for starters, the sum of expressions:
Fine! this expression is exactly equal to twice the product. Then, in order to get the square of the sum in brackets, you need to add and subtract the double product:
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