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

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

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

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

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

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

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

We introduced the equation above in § 7. First, we recall what a rational expression is. This is an algebraic expression made up 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 a somewhat broader interpretation of the 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 relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.

1) Let's transform the equation to the form

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

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

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

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

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

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

Answer: 4, - 1.

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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.

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: .

Solution:

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: .

Solution

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

T. Kosyakova,
school N№ 80, Krasnodar

Solution of quadratic and fractional-rational equations containing parameters

Lesson 4

Lesson topic:

The purpose of the lesson: to form the ability to solve fractional-rational equations containing parameters.

Lesson type: introduction of new material.

1. (Oral.) Solve the equations:

Example 1. Solve the Equation

Solution.

Find invalid values a:

Answer. If a if a = – 19 , then there are no roots.

Example 2. Solve the Equation

Solution.

Find invalid parameter values a :

10 – a = 5, a = 5;

10 – a = a, a = 5.

Answer. If a a = 5 a № 5 , then x=10– a .

Example 3. At what values ​​of the parameter b the equation It has:

a) two roots b) the only root?

Solution.

1) Find invalid parameter values b :

x= b, b 2 (b 2 – 1) – 2b 3 + b 2 = 0, b 4 – 2b 3 = 0,
b= 0 or b = 2;
x = 2, 4( b 2 – 1) – 4b 2 + b 2 = 0, b 2 – 4 = 0, (b – 2)(b + 2) = 0,
b= 2 or b = – 2.

2) Solve the equation x 2 ( b 2 – 1) – 2b 2x+ b 2 = 0:

D=4 b 4 – 4b 2 (b 2 – 1), D = 4 b 2 .

a)

Excluding invalid parameter values b , we get that the equation has two roots, if b № – 2, b № – 1, b № 0, b № 1, b № 2 .

b) 4b 2 = 0, b = 0, but this is an invalid parameter value b ; if b 2 –1=0 , i.e. b=1 or.

Answer: a) if b № –2 , b № –1, b № 0, b № 1, b № 2 , then two roots; b) if b=1 or b=-1 , then the only root.

Independent work

Option 1

Solve the equations:

Option 2

Solve the equations:

Answers

IN 1. what if a=3 , then there are no roots; if b) if if a № 2 , then there are no roots.

IN 2. If a a=2 , then there are no roots; if a=0 , then there are no roots; if
b) if a=– 1 , then the equation loses its meaning; if then there are no roots;
if

Homework assignment.

Solve the equations:

Answers: a) If a № –2 , then x= a ; if a=–2 , then there are no solutions; b) if a № –2 , then x=2; if a=–2 , then there are no solutions; c) if a=–2 , then x- any number other than 3 ; if a № –2 , then x=2; d) if a=–8 , then there are no roots; if a=2 , then there are no roots; if

Lesson 5

Lesson topic:"Solution of Fractional-Rational Equations Containing Parameters".

Lesson Objectives:

learning to solve equations with a non-standard condition;
conscious assimilation by students of algebraic concepts and relationships between them.

Lesson type: systematization and generalization.

Checking homework.

Example 1. Solve the Equation

a) relative to x; b) relative to y.

Solution.

a) Find invalid values y: y=0, x=y, y2=y2 –2y,

y=0– invalid parameter value y.

If a y№ 0 , then x=y-2; if y=0, then the equation loses its meaning.

b) Find invalid parameter values x: y=x, 2x–x 2 +x 2 =0, x=0– invalid parameter value x; y(2+x-y)=0, y=0 or y=2+x;

y=0 does not satisfy the condition y(y–x)№ 0 .

Answer: a) if y=0, then the equation loses its meaning; if y№ 0 , then x=y-2; b) if x=0 x№ 0 , then y=2+x .

Example 2. For what integer values ​​of the parameter a are the roots of the equation belong to the interval

D = (3 a + 2) 2 – 4a(a+ 1) 2 = 9 a 2 + 12a + 4 – 8a 2 – 8a,

D = ( a + 2) 2 .

If a a № 0 or a № – 1 , then

Answer: 5 .

Example 3. Find relatively x entire solutions of the equation

Answer. If a y=0, then the equation does not make sense; if y=–1, then x- any integer other than zero; if y# 0, y# – 1, then there are no solutions.

Example 4 Solve the Equation with parameters a and b .

If a a№ – b , then

Answer. If a a= 0 or b= 0 , then the equation loses its meaning; if a№ 0,b№ 0, a=-b , then x- any number other than zero; if a№ 0,b№ 0,a№ -b then x=-a, x=-b .

Example 5. Prove that for any non-zero value of the parameter n, the equation has a single root equal to – n .

Solution.

i.e. x=-n, which was to be proved.

Homework assignment.

1. Find entire solutions of the equation

2. At what values ​​of the parameter c the equation It has:
a) two roots b) the only root?

3. Find all integer roots of the equation if a O N .

4. Solve the equation 3xy - 5x + 5y = 7: a) relatively y; b) relatively x .

1. The equation is satisfied by any integer equal values ​​of x and y other than zero.
2. a) When
b) at or
3. – 12; – 9; 0 .
4. a) If then there are no roots; if
b) if then there are no roots; if

Test

Option 1

1. Determine the type of equation 7c(c + 3)x 2 +(c–2)x–8=0 at: a) c=-3; b) c=2 ; in) c=4 .

2. Solve the equations: a) x 2 –bx=0; b) cx 2 –6x+1=0; in)

3. Solve the equation 3x-xy-2y=1:

a) relatively x ;
b) relatively y .

nx 2 - 26x + n \u003d 0, knowing that the parameter n takes only integer values.

5. For what values ​​of b does the equation It has:

a) two roots
b) the only root?

Option 2

1. Determine the type of equation 5c(c + 4)x 2 +(c–7)x+7=0 at: a) c=-4 ; b) c=7 ; in) c=1 .

2. Solve the equations: a) y 2 +cy=0 ; b) ny2 –8y+2=0; in)

3. Solve the equation 6x-xy+2y=5:

a) relatively x ;
b) relatively y .

4. Find the integer roots of the equation nx 2 -22x+2n=0 , knowing that the parameter n takes only integer values.

5. For what values ​​of the parameter a the equation It has:

a) two roots
b) the only root?

Answers

IN 1. 1. a) Linear equation;
b) incomplete quadratic equation; c) a quadratic equation.
2. a) If b=0, then x=0; if b#0, then x=0, x=b;
b) if cО (9;+Ґ ), then there are no roots;
c) if a=–4 , then the equation loses its meaning; if a№ –4 , then x=- a .
3. a) If y=3, then there are no roots; if);
b) a=–3, a=1.

Additional tasks

Solve the equations:

Literature

1. Golubev V.I., Goldman A.M., Dorofeev G.V. About the parameters from the very beginning. - Tutor, No. 2/1991, p. 3–13.
2. Gronshtein P.I., Polonsky V.B., Yakir M.S. Necessary conditions in tasks with parameters. – Kvant, No. 11/1991, p. 44–49.
3. Dorofeev G.V., Zatakavai V.V. Solving problems containing parameters. Part 2. - M., Perspective, 1990, p. 2–38.
4. Tynyakin S.A. Five hundred and fourteen tasks with parameters. - Volgograd, 1991.
5. Yastrebinetsky G.A. Tasks with parameters. - M., Education, 1986.