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Systems of fractionally rational equations. The simplest rational equations. Examples

We have already learned 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 a solution algorithm rational equations:

1. Transfer all terms to left side to get 0 on the right side.

2. Transform and simplify the left side, bring all fractions to 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 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 " Public lesson" ().
2. School.xvatit.com().
3. Rudocs.exdat.com().

Homework

\(\bullet\) A rational equation is an equation expressed as \[\dfrac(P(x))(Q(x))=0\] where \(P(x), \ Q(x)\) - polynomials (the sum of "xes" in various degrees, multiplied by various numbers).
The expression on the left side of the equation is called the rational expression.
ODZ (region allowed values) of a rational equation are all values ​​\(x\) for which the denominator does NOT vanish, i.e. \(Q(x)\ne 0\) .
\(\bullet\) 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 are all \(x\) such that \(x\ne 3\) (they 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 zero, while the other does not lose its meaning, therefore, the equation \(f(x)\cdot g(x)=0\) is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODV equations) \end(cases)\] 2) The 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 the system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]\(\bullet\) Let's look at some examples.

1) Solve the equation \(x+1=\dfrac 2x\) . Let's find the ODZ of this equation - this is \(x\ne 0\) (since \(x\) is in the denominator).
So, the ODZ can be written as follows: .
Let's transfer all the terms into one part and reduce 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 us find the ODZ of this equation. We see that the only value \(x\) for which the left side does not make sense is \(x=0\) . So the OD can be written as follows: \(x\in (-\infty;0)\cup(0;+\infty)\).
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 \(x=0\) in the original equation, then it will not make sense, because the expression \(\dfrac 40\) is not defined.
So 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 \(4x^2-1\ne 0\) , whence \((2x-1)(2x+1)\ne 0\) , i.e. \(x\ne -\frac12; \frac12\) .
We transfer all the terms to the left side and reduce 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 through a semicolon in curly braces, as shown in the previous examples.

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Smirnova Anastasia Yurievna

Lesson type: lesson learning new material.

Form of organization learning activities : frontal, individual.

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

Lesson objectives.

Tutorial:

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

Developing:

• create conditions for the formation of skills to apply the acquired knowledge;
• to promote the development of students' cognitive interest in 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.

Nurturing:

• education of cognitive interest in the subject;
• education of independence in solving educational problems;
• education of will and perseverance to achieve the final results.

Equipment: textbook, blackboard, crayons.

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

On this topic five hours are allotted. This lesson is the first. The main thing is to study the algorithm for solving fractional rational equations and work out 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 could,
Smile, good luck to everyone
No matter what problems
Smiled at each other, created good mood and started work.

Equations are written on the blackboard, 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 today in the lesson? Formulate the topic of the lesson. So, we open notebooks and write down the topic of the lesson “Solution of fractional rational equations”.

2. Actualization of knowledge. Frontal survey, oral work with the class.

And now we will repeat the main theoretical material that we 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 equation #1 called? ( Linear.) Method of solution linear equations. (Move everything with the unknown to the left side of the equation, all numbers to the right. Bring like terms. Find the unknown multiplier).
3. What is Equation 3 called? ( Square.) Methods for solving quadratic equations. (P about formulas)
4. What is a proportion? ( Equality of two relations.) The main property of proportion. ( If the proportion is true, then the product of its extreme terms is equal to the product of the middle terms.)
5. What properties are used to solve equations? ( 1. If in the equation we transfer the term from one part to another, changing its sign, then we get an equation equivalent to the given one. 2. If both parts of the equation are multiplied or divided by the same non-zero number, then an equation will be obtained that is equivalent to the given.)
6. When is a fraction equal to zero? ( A fraction is zero when the numerator is zero and the denominator is non-zero.)

3. Explanation of new material.

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

Answer: 10.

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

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

x 2 -4x-2x + 8 \u003d x 2 + 3x + 2x + 6

x 2 -6x-x 2 -5x \u003d 6-8

Solve equation No. 4 in 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 consider the solution of equations of the type of 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 met the concept of an extraneous root, it is really 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. 5.6? ( In equations No. 2 and 4 in the denominator of the number, No. 5-6 - expressions with a variable.)
• What is the root of the equation? ( The value of the variable at which the equation becomes a true equality.)
• How to find out if a number is the root of an equation? ( Make a check.)

When doing a test, 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 eliminates 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 themselves formulate the algorithm.

Algorithm for solving fractional rational equations:

1. Move everything to the left.
2. Bring fractions to a common denominator.
3. Make up a system: a fraction is zero when the numerator is zero and the denominator is not zero.
4. Solve the equation.
5. Check inequality to exclude extraneous roots.
6. Write down the answer.

4. Primary comprehension of new material.

Work in pairs. Students choose how to solve the equation on their own, depending on the type of equation. Tasks from the textbook "Algebra 8", Yu.N. Makarychev, 2007: No. 600(b, c); No. 601(a, e). The teacher controls the performance of the task, answers the questions that have arisen, and provides assistance to poorly 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. Statement of homework.

1. Read item 25 from the textbook, analyze examples 1-3.
2. Learn the algorithm for solving fractional rational equations.
3. Solve in notebooks No. 600 (d, e); 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 fractional rational equations are solved, what should be kept in mind? What is the "cunning" of fractional rational equations?

Thank you all, the lesson is over.

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

Additional materials
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Teaching aids and simulators in the online store "Integral" for grade 8
Manual for the textbook Makarychev Yu.N. Manual for the textbook Mordkovich A.G.

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. The concept of rational equations is in many ways similar to the concept of 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 1
Solve 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 zero if and only if the numerator of the fraction is zero and the denominator is non-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. Move all expressions contained in the equation to the left of 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: $x^2+x-2=0$.
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.
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. $(x+2)(x+3)(x+4)(x+5)=3$.