Solving rational equations examples of solutions. The simplest rational equations

$\bullet$ A rational equation is an equation represented in the form \[\dfrac(P(x))(Q(x))=0\] where $P(x), \Q(x)$ - polynomials (the sum of “X’s” in various powers, multiplied by various numbers).
The expression on the left side of the equation is called a rational expression.
The EA (range of acceptable values) of a rational equation is all the values of $x$ at which the denominator does NOT vanish, that is, $Q(x)\ne 0$ .
$\bullet$ For example, equations \[\dfrac(x+2)(x-3)=0,\qquad \dfrac 2(x^2-1)=3, \qquad x^5-3x=2\] are rational equations.
In the first ODZ equation– these are all $x$ such that $x\ne 3$ (write $x\in (-\infty;3)\cup(3;+\infty)$); in the second equation – these are all $x$ such that $x\ne -1; x\ne 1$ (write $x\in (-\infty;-1)\cup(-1;1)\cup(1;+\infty)$); and in the third equation there are no restrictions on the ODZ, that is, the ODZ is all $x$ (they write $x\in\mathbb(R)$). $\bullet$ Theorems:
1) The product of two factors is equal to zero if and only if one of them equal to zero, and the other does not lose meaning, therefore, the equation $f(x)\cdot g(x)=0$ is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODZ equations)\end(cases)\] 2) A fraction is equal to zero if and only if the numerator is equal to zero and the denominator is not equal to zero, therefore, the equation $\dfrac(f(x))(g(x))=0$ is equivalent to a system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]$\bullet$ Let's look at a few examples.

1) Solve the equation $x+1=\dfrac 2x$ . Let's find ODZ given equation is $x\ne 0$ (since $x$ is in the denominator).
This means that the ODZ can be written as follows: .
Let's move all the terms into one part and bring them to a common denominator: \[\dfrac((x+1)\cdot x)x-\dfrac 2x=0\quad\Leftrightarrow\quad \dfrac(x^2+x-2)x=0\quad\Leftrightarrow\quad \begin( cases) x^2+x-2=0\\x\ne 0\end(cases)\] The solution to the first equation of the system will be $x=-2, x=1$ . We see that both roots are non-zero. Therefore, the answer is: $x\in \(-2;1$\) .

2) Solve the equation $\left(\dfrac4x - 2\right)\cdot (x^2-x)=0$. Let's find the ODZ of this equation. We see that the only value of $x$ for which the left side does not make sense is $x=0$ . So, the ODZ can be written like this: $x\in (-\infty;0)\cup(0;+\infty)$.
Thus, this equation is equivalent to the system:

\[\begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x-2=0\\ &x^2-x=0 \end(aligned) \end(gathered) \right. \\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x=2\\ &x(x-1)= 0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &x =2\\ &x=1\\ &x=0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \left[ \begin(gathered) \begin(aligned) &x=2\\ &x=1 \end(aligned) \end(gathered) \right.\] Indeed, despite the fact that $x=0$ is the root of the second factor, if you substitute $x=0$ into the original equation, then it will not make sense, because expression $\dfrac 40$ is not defined.
Thus, the solution to this equation is $x\in \(1;2$\) .

3) Solve the equation \[\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1)\] In our equation $4x^2-1\ne 0$ , from which $(2x-1)(2x+1)\ne 0$ , that is, $x\ne -\frac12; \frac12$ .
Let's move all terms to left side and bring it to a common denominator:

$\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1) \quad \Leftrightarrow \quad \dfrac(x^2+4x- 3+x+x^2)(4x^2-1)=0\quad \Leftrightarrow \quad \dfrac(2x^2+5x-3)(4x^2-1)=0 \quad \Leftrightarrow$

$\Leftrightarrow \quad \begin(cases) 2x^2+5x-3=0\\ 4x^2-1\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) (2x-1 )(x+3)=0\\ (2x-1)(2x+1)\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered) \begin( aligned) &x=\dfrac12\\ &x=-3 \end(aligned)\end(gathered) \right.\\ x\ne \dfrac 12\\ x\ne -\dfrac 12 \end(cases) \quad \ Leftrightarrow \quad x=-3$

Answer: $x\in \(-3$\) .

Comment. If the answer consists of a finite set of numbers, then they can be written separated by semicolons in curly braces, as shown in the previous examples.

Problems that require solving rational equations are encountered every year in the Unified State Examination in mathematics, so when preparing to pass the certification test, graduates should definitely repeat the theory on this topic on their own. Graduates taking both the basic and profile level exam. Having mastered the theory and dealt with practical exercises on the topic “Rational Equations”, students will be able to solve problems with any number of actions and count on receiving competitive scores based on the results of passing the Unified State Exam.

How to prepare for the exam using the Shkolkovo educational portal?

Sometimes you can find a source that fully presents the basic theory for solving mathematical problems turns out to be quite difficult. The textbook may simply not be at hand. And find necessary formulas sometimes it can be quite difficult even on the Internet.

The Shkolkovo educational portal will relieve you of the need to search the required material and will help you prepare well for passing the certification test.

All necessary theory on the topic “Rational Equations” our specialists prepared and presented to the maximum accessible form. After studying the information presented, students will be able to fill gaps in knowledge.

For successful preparation To Unified State Examination for graduates it is necessary not only to brush up on the basic theoretical material on the topic “Rational Equations”, but to practice completing tasks on specific examples. A large selection of tasks is presented in the “Catalogue” section.

For each exercise on the site, our experts have written a solution algorithm and indicated the correct answer. Students can practice solving problems varying degrees difficulties depending on the level of training. The list of tasks in the corresponding section is constantly supplemented and updated.

Study theoretical material and hone problem-solving skills on the topic “Rational Equations”, similar to those included in Unified State Exam tests, can be done online. If necessary, any of the presented tasks can be added to the “Favorites” section. Repeating again basic theory on the topic “Rational Equations”, a high school student will be able to return to the problem in the future to discuss the progress of its solution with the teacher in an algebra lesson.

Presentation and lesson on the topic: "Rational equations. Algorithm and examples of solving rational equations"

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Introduction to Irrational Equations

Guys, we learned how to solve quadratic equations. But mathematics is not limited to them only. Today we will learn how to solve rational equations. Concept rational equations is 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 the operations of addition, subtraction, multiplication, division and raising to an integer power are present.

Let $r(x)$ be rational expression . Such an expression can be a simple polynomial in the variable $x$ or a ratio of polynomials (a division operation 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.

Let's look at examples of solving rational equations.

Example 1.
Solve the equation: $\frac(5x-3)(x-3)=\frac(2x-3)(x)$.

Solution.
Let's move all the expressions to the left side: $\frac(5x-3)(x-3)-\frac(2x-3)(x)=0$.
If the left side of the equation were represented ordinary numbers, 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 is zero and the denominator is non-zero. Then we 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 coincide. So we write down both roots of the numerator in the answer.
Answer: $x=1$ or $x=-3$.

If suddenly one of the roots of the numerator coincides with the root of the denominator, then it should be excluded. Such roots are called extraneous!

Algorithm for solving rational equations:

1. Transfer all expressions contained in the equation 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 coincide 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.
Let's 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 $x=1$ coincides with the root of the numerator, then we do not write it down in the answer.
Answer: $x=-1$.

It is convenient to solve rational equations using the change of variables method. Let's demonstrate this.

Example 3.
Solve the equation: $x^4+12x^2-64=0$.

Solution.
Let's introduce the replacement: $t=x^2$.
Then our equation will take the form:
$t^2+12t-64=0$ - ordinary quadratic equation.
$t_(1,2)=\frac(-12±\sqrt(12^2-4*(-64)))(2)=\frac(-12±20)(2)=-16; $4.
Let's introduce the reverse substitution: $x^2=4$ or $x^2=-16$.
The roots of the first equation are a pair of numbers $x=±2$. The second thing is that it 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 proceed 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 coincide.
Let's 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: $x^2+x-2=0$.
The roots of this equation will be the 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.
Let's introduce the 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'll 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: $x=\frac(-3±\sqrt(5))(2)$, $x=1$.

Problems to solve independently

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 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 coincide with invalid values variables that were obtained by 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. Tutorial for educational institutions. - M.: Education, 2006.

Festival pedagogical ideas "Public lesson" ().
School.xvatit.com ().
Rudocs.exdat.com ().

Homework

Smirnova Anastasia Yurievna

Lesson type: lesson of learning new material.

Form of organization educational activities : frontal, individual.

The purpose of the lesson: to introduce a new type of equations - fractional rational equations, to give an idea of the algorithm for solving fractional rational equations.

Lesson objectives.

Educational:

formation of the concept of a fractional rational equation;
consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero;
teach solving fractional rational equations using an algorithm.

Developmental:

create conditions for developing skills in applying acquired knowledge;
promote development cognitive interest students to the subject;
developing students’ ability to analyze, compare and draw conclusions;
development of skills of mutual control and self-control, attention, memory, oral and writing, independence.

Educating:

fostering cognitive interest in the subject;
fostering independence in decision-making educational tasks;
nurturing will and perseverance to achieve final results.

Equipment: textbook, blackboard, crayons.

Textbook "Algebra 8". Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, S.B. Suvorova, edited by S.A. Telyakovsky. Moscow "Enlightenment". 2010

On this topic five hours are allotted. This is the first lesson. The main thing is to study the algorithm for solving fractional rational equations and practice this algorithm in exercises.

During the classes

1. Organizational moment.

Hello guys! Today I would like to start our lesson with a quatrain:
To make life easier for everyone,
What would be decided, what would be possible,
Smile, good luck to everyone,
So that there are no problems,
We smiled at each other and created good mood and started work.

There are equations written on the board, look at them carefully. Can you solve all of these equations? Which ones are not and why?

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

2. Updating knowledge. Frontal survey, oral work with class.

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

What is an equation? ( Equality with a variable or variables.)
What is the name of equation number 1? ( Linear.) Solution linear equations. (Move everything with the unknown to the left side of the equation, all numbers to the right. Lead similar terms. Find unknown factor).
What is the name of equation number 3? ( Square.) Solutions quadratic equations. (P about formulas)
What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)
What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)
When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)

3. Explanation of new material.

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

Answer: 10.

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

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

x 2 -4x-2x+8 = x 2 +3x+2x+6

x 2 -6x-x 2 -5x = 6-8

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

Answer: 1,5.

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

x 2 -7x+12 = 0

D=1›0, x 1 =3, x 2 =4.

Answer: 3;4.

We will look at solving equations like equation No. 7 in the following lessons.

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

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

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

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

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

Algorithm for solving fractional rational equations:

Move everything to the left side.
Reduce fractions to a common denominator.
Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.
Solve the equation.
Check inequality to exclude extraneous roots.
Write down the answer.

4. Initial comprehension of new material.

Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, Yu.N. Makarychev, 2007: No. 600(b,c); No. 601(a,e). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.

b) 2 - extraneous root. Answer: 3.

c) 2 - extraneous root. Answer: 1.5.

a) Answer: -12.5.

5. Setting homework.

Read paragraph 25 from the textbook, analyze examples 1-3.
Learn an algorithm for solving fractional rational equations.
Solve in notebooks No. 600 (d, d); No. 601(g,h).

6. Summing up the lesson.

So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations different ways. Regardless of how you solve fractional rational equations, what should you keep in mind? What is the “cunning” of fractional rational equations?

Thanks everyone, lesson is over.

"Solving fractional rational equations"

Lesson objectives:

Educational:

formation of the concept of fractional rational equations; consider various ways to solve fractional rational equations; consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero; teach solving fractional rational equations using an algorithm; checking the level of mastery of the topic by conducting a test.

Developmental:

mental operations

critical thinking

Educating:

fostering cognitive interest in the subject; fostering independence in solving educational problems; nurturing will and perseverance to achieve final results.

Lesson type: lesson - explanation of new material.

During the classes

1. Organizational moment.

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

2. Updating knowledge. Frontal survey, oral work with the class.

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

1. What is an equation? ( Equality with a variable or variables.)

2. What is the name of equation No. 1? ( Linear.) A method for solving linear equations. ( Move everything with the unknown to the left side of the equation, all numbers to the right. Give similar terms. Find unknown factor).

3. What is the name of equation No. 3? ( Square.) Methods for solving quadratic equations. ( Selection full square, by formulas, using Vieta’s theorem and its consequences.)

4. What is proportion? ( Equality of two ratios.) The main property of proportion. ( If the proportion is correct, then the product of its extreme terms is equal to the product of the middle terms.)

5. What properties are used when solving equations? ( 1. If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one. 2. If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.)

6. When does a fraction equal zero? ( A fraction is equal to zero when the numerator is zero and the denominator is not zero..)

3. Explanation of new material.

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

Answer: 10.

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

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

x2-4x-2x+8 = x2+3x+2x+6

x2-6x-x2-5x = 6-8

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

Answer: 1,5.

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

D=1›0, x1=3, x2=4.

Answer: 3;4.

Now try to solve equation number 7 using one of the following methods.


(x2-2x-5)x(x-5)=x(x-5)(x+5)
(x2-2x-5)x(x-5)-x(x-5)(x+5)=0
x(x-5)(x2-2x-5-(x+5))=0			x2-2x-5-x-5=0
x(x-5)(x2-3x-10)=0
x=0 x-5=0 x2-3x-10=0
x1=0 x2=5 D=49

Answer: 0;5;-2.			Answer: 5;-2.

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

In equations No. 2 and 4 there are numbers in the denominator, No. 5-7 are expressions with a variable

The value of the variable at which the equation becomes true

Make a check

x2-3x-10=0, D=49, x1=5, x2=-2.

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

If x=-2, then x(x-5)≠0.

Answer: -2.

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

Algorithm for solving fractional rational equations:

1. Move everything to the left side.

2. Reduce fractions to a common denominator.

3. Create a system: a fraction is equal to zero when the numerator is equal to zero and the denominator is not equal to zero.

4. Solve the equation.

5. Check the inequality to exclude extraneous roots.

6. Write down the answer.

Discussion: how to formalize the solution if the basic property of proportion is used and both sides of the equation are multiplied by common denominator. (Add to the solution: exclude from its roots those that make the common denominator vanish).

4. Initial comprehension of new material.

Work in pairs. Students choose how to solve the equation themselves depending on the type of equation. Assignments from the textbook “Algebra 8”, 2007: No. 000 (b, c, i); No. 000(a, d, g). The teacher monitors the completion of the task, answers any questions that arise, and provides assistance to low-performing students. Self-test: answers are written on the board.

b) 2 – extraneous root. Answer: 3.

c) 2 – extraneous root. Answer: 1.5.

a) Answer: -12.5.

g) Answer: 1;1.5.

5. Setting homework.

2. Learn the algorithm for solving fractional rational equations.

3. Solve in notebooks No. 000 (a, d, e); No. 000(g, h).

4. Try to solve No. 000(a) (optional).

6. Completing a control task on the topic studied.

The work is done on pieces of paper.

Example task:

A) Which of the equations are fractional rational?

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

Q) Is the number -3 the root of equation number 6?

D) Solve equation No. 7.

Assessment criteria for the assignment:

“5” is given if the student completed more than 90% of the task correctly. “4” - 75%-89% “3” - 50%-74% “2” is given to a student who has completed less than 50% of the task. A rating of 2 is not given in the journal, 3 is optional.

7. Reflection.

On the independent work sheets, write:

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

8. Summing up the lesson.

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

Which method of solving fractional rational equations, in your opinion, is easier, more accessible, and more rational? Regardless of the method for solving fractional rational equations, what should you remember? What is the “cunning” of fractional rational equations?

Thanks everyone, lesson is over.