Examples of solving fractional rational equations. Video lesson "Rational Equations

\(\bullet\) 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.
The ODV (range of acceptable values) of a rational equation is 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 equation, the ODZ is 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 ODZ given equation is \(x\ne 0\) (because \(x\) 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:

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

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

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.

Tasks to be solved rational equations, in the Unified State Examination in mathematics they meet every year, therefore, in preparation for passing the certification test, graduates should definitely repeat the theory on this topic on their own. To be able to cope with such tasks, graduates who pass both the basic and the profile level of the exam must necessarily. 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 expect to receive competitive points based on the results of passing the exam.

How to prepare for the exam with the educational portal "Shkolkovo"?

Sometimes it is quite difficult to find a source in which the basic theory for solving mathematical problems is fully presented. The textbook may simply not be at hand. And sometimes it is quite difficult to find the necessary formulas even on the Internet.

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All the necessary theory on the topic "Rational Equations" was prepared by our specialists and presented in the most accessible form. By studying the information presented, students will be able to fill in the gaps in knowledge.

For successful preparation for the exam, graduates need not only to brush up on the basic theoretical material on the topic "Rational Equations", but to practice doing assignments on concrete examples. Big selection tasks is presented in the "Catalogue" section.

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

Study the theoretical material and hone the skills of solving problems on the topic "Rational Equations", similar to those included in USE tests, you can online. If necessary, any of the presented tasks can be added to the "Favorites" section. Having once again repeated the basic theory on the topic "Rational Equations", the 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 the algebra lesson.

Lesson Objectives:

Tutorial:

• formation of the concept of fractional rational equations;
• to consider various ways of solving fractional rational equations;
• 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;
• checking the level of assimilation of the topic by conducting test work.

Developing:

• development of the ability to correctly operate with the acquired knowledge, to think logically;
• development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization;
• development of initiative, the ability to make decisions, not to stop there;
• development critical thinking;
• development of research skills.

Nurturing:

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

Lesson type: lesson - explanation of new material.

During the classes

1. Organizational moment.

Hello guys! 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 parts 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 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. ( Selection of the full square, by formulas, using the Vieta theorem and its consequences.)
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.

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)

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.

Now try to solve equation #7 in one of the ways.

(x 2 -2x-5)x(x-5)=x(x-5)(x+5)
(x 2 -2x-5)x(x-5)-x(x-5)(x+5)=0			x 2 -2x-5=x+5
x(x-5)(x 2 -2x-5-(x+5))=0			x 2 -2x-5-x-5=0
x(x-5)(x 2 -3x-10)=0
x=0 x-5=0 x 2 -3x-10=0
x 1 \u003d 0 x 2 \u003d 5 D \u003d 49
x 3 \u003d 5 x 4 \u003d -2			x 3 \u003d 5 x 4 \u003d -2
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?

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. 5,6,7? ( In equations No. 2 and 4 in the denominator of the number, No. 5-7 - 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.

x 2 -3x-10=0, D=49, x 1 =5, x 2 = -2.

If x=5, then x(x-5)=0, so 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 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.

Discussion: how to formalize the solution if the basic property of proportion is used and the multiplication of both sides of the equation by a common denominator. (Supplement the solution: exclude from its roots those that turn the common denominator to zero).

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, i); No. 601(a, e, g). 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 is an extraneous root. Answer:3.

c) 2 is an extraneous root. Answer: 1.5.

a) Answer: -12.5.

g) Answer: 1; 1.5.

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 (a, d, e); No. 601 (g, h).
4. Try to solve #696(a) (optional).

6. Fulfillment of the control task on the studied topic.

The work is done on sheets.

Job example:

A) Which of the equations are fractional rational?

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

Q) Is the number -3 the root of Equation #6?

D) Solve equation No. 7.

Task evaluation criteria:

• "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 completed less than 50% of the task.
• Grade 2 is not put in the journal, 3 is optional.

7. Reflection.

On the leaflets with independent work, put:

• 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 how to solve these equations different ways, tested their knowledge with the help of training independent work. You will learn the results of independent work in the next lesson, at home you will have the opportunity to consolidate the knowledge gained.

What method of solving fractional rational equations, in your opinion, is easier, more accessible, more rational? Regardless of the method of solving fractional rational equations, what should not be forgotten? What is the "cunning" of fractional rational equations?

Thank you all, the lesson is over.

Solution fractional rational equations

Help Guide

Rational equations are equations in which both the left and right sides are rational expressions.

(Recall that rational expressions are integers and fractional expressions without radicals, including operations of addition, subtraction, multiplication or division - for example: 6x; (m – n)2; x/3y etc.)

Fractional-rational equations, as a rule, are reduced to the form:

Where P(x) And Q(x) are polynomials.

To solve such equations, multiply both sides of the equation by Q(x), which can lead to the appearance of extraneous roots. Therefore, when solving fractional rational equations, it is necessary to check the found roots.

A rational equation is called an integer, or algebraic, if it does not have a division by an expression containing a variable.

Examples of a whole rational equation:

5x - 10 = 3(10 - x)

3x
-=2x-10
4

If in a rational equation there is a division by an expression containing the variable (x), then the equation is called fractional rational.

An example of a fractional rational equation:

15
x + - = 5x - 17
x

Fractional rational equations are usually solved as follows:

1) find a common denominator of fractions and multiply both parts of the equation by it;

2) solve the resulting whole equation;

3) exclude from its roots those that turn the common denominator of the fractions to zero.

Examples of solving integer and fractional rational equations.

Example 1. Solve the whole equation

x – 1 2x 5x
-- + -- = --.
2 3 6

Solution:

Finding the lowest common denominator. This is 6. Divide 6 by the denominator and multiply the result by the numerator of each fraction. We get an equation equivalent to this one:

3(x - 1) + 4x 5x
------ = --
6 6

Since the left and right sides same denominator, it can be omitted. Then we have a simpler equation:

3(x - 1) + 4x = 5x.

We solve it by opening brackets and reducing like terms:

3x - 3 + 4x = 5x

3x + 4x - 5x = 3

Example solved.

Example 2. Solve a fractional rational equation

x – 3 1 x + 5
-- + - = ---.
x - 5 x x(x - 5)

We find a common denominator. This is x(x - 5). So:

x 2 – 3x x – 5 x + 5
--- + --- = ---
x(x - 5) x(x - 5) x(x - 5)

Now we get rid of the denominator again, since it is the same for all expressions. We reduce like terms, equate the equation to zero and get quadratic equation:

x 2 - 3x + x - 5 = x + 5

x 2 - 3x + x - 5 - x - 5 = 0

x 2 - 3x - 10 = 0.

Having solved the quadratic equation, we find its roots: -2 and 5.

Let's check if these numbers are the roots of the original equation.

For x = –2, the common denominator x(x – 5) does not vanish. So -2 is the root of the original equation.

At x = 5, the common denominator vanishes, and two of the three expressions lose their meaning. So the number 5 is not the root of the original equation.

Answer: x = -2

More examples

Example 1

x 1 \u003d 6, x 2 \u003d - 2.2.

Answer: -2.2; 6.

Example 2

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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