Solution of fractionally rational equations. Algorithm for solving rational equations

Today we will figure out how to solve fractional rational equations.

Let's see: from the equations

(1) 2x + 5 = 3(8 - x),

(3)

(4)

fractional rational equations are only (2) and (4), while (1) and (3) are entire equations.

I propose to solve equation (4), and then formulate the rule.

Since the equation is fractional, we need to find a common denominator. In this equation, this expression is 6 (x - 12) (x - 6). Then we multiply both sides of the equation by a common denominator:

After reduction, we get the whole equation:

6 (x - 6) 2 - 6 (x - 12) 2 \u003d 5 (x - 12) (x - 6).

Having solved this equation, it is necessary to check whether the obtained roots turn the denominators of the fractions in the original equation to zero.

Expanding the brackets:
6x 2 - 72x + 216 - 6x 2 + 144x - 864 = 5x 2 - 90x + 360, we simplify the equation: 5x 2 - 162x + 1008 = 0.

Finding the roots of the equation
D=6084, √D=78,
x 1 = (162 - 78) / 10 = 84/10 = 8.4 and x 2 = (162 + 78) / 10 = 240/10 = 24.

At x = 8.4 and 24, the common denominator is 6(x - 12)(x - 6) ≠ 0, which means that these numbers are the roots of equation (4).

Answer: 8,4; 24.

Solving the proposed equation, we arrive at the following provisions:

1) We find a common denominator.

2) Multiply both sides of the equation by a common denominator.

3) We solve the resulting whole equation.

4) We check which of the roots turn the common denominator to zero and exclude them from the solution.

Let us now look at an example of how the resulting positions work.

Solve the equation:

1) Common denominator: x 2 - 1

2) We multiply both parts of the equation by a common denominator, we get the whole equation: 6 - 2 (x + 1) \u003d 2 (x 2 - 1) - (x + 4) (x - 1)

3) We solve the equation: 6 - 2x - 2 \u003d 2x 2 - 2 - x 2 - 4x + x + 4

x 2 - x - 2 = 0

x 1 = - 1 and x 2 = 2

4) When x \u003d -1, the common denominator x 2 - 1 \u003d 0. The number -1 is not a root.

For x \u003d 2, the common denominator is x 2 - 1 ≠ 0. The number 2 is the root of the equation.

Answer: 2.

As you can see, our provisions work. Don't be afraid, you will succeed! The most important thing find the common denominator correctly and do the transformations carefully. We hope that when solving fractional rational equations, you will always get the correct answers. If you have any questions or want to practice solving such equations, sign up for lessons with the author of this article, tutor J.

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The least common denominator is used to simplify this equation. This method is applicable in the case when it is impossible to write this equation with one rational expression on each side of the equation (and use the cross-multiplication method). This method is used when a rational equation with three or more fractions is given (in the case of two fractions it is better to use crosswise multiplication).

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).

Look for "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 two 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.

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Solution of fractional rational equations

If you are an eighth grade student, and suddenly it happened that you missed a lesson or missed what the teacher was talking about, this article is for you!

To begin with, let's figure out what it is - fractional rational equations? In any textbook there is such a definition: A fractional-rational equation is an equation of the form\(fxg(x)=0\) .

And of course, this definition doesn't tell you anything. Then I give examples, and you try to identify a pattern, find something in common.

\(((-2x-4)\over (x^2-4))=((x+5)\over (x-2))\)\(((3x^2-6)\over 2(x+1)) =x-1\)\((x\over x-2 ) + (8\over(4-x^2)) - (1\over x+2)=0\)

And these equations are not fractionally rational:

\(3x^2+x-25=0 \) \(((2-x)\over (2))+((3x\over 5))=4\)\(((2x-1)\over 2)+(5x\over6)-(1-x\over 3)=3x-2\)

The last two equations are definitely not fractionally rational, despite the fact that they consist of fractions. But the most important thing is that there is no variable (letter) in the denominator. But in a fractional rational equation, there is always a variable in the denominator.

So, after you have correctly determined what exactly the equation is before you, we will begin to solve it. The first thing to do is indicated by three capital letters,O.D.Z.What do these letters mean?O blast D acceptable Wideas. What this means in the science of mathematics, I will not explain now, our goal is to learn how to solve equations, and not repeat the topic “Algebraic fractions”. But for our purpose, this means the following: we take the denominator or denominators of our fractions, write them out separately and note that they are not equal to zero.

If we use our equations as an example\(((-2x-4)\over x^2-4)=(x+5\over x-2)\), we do this:

ODZ: \(x^2-4≠0 \)

\(x-2≠0 \)

\((3x^2-6\over 2(x+1)) =x-1 \)

ODZ: \(x+1≠0\)

Why didn't they specify a factor of 2? It is so clear that 2≠0

\((x\over x-2)+(8\over 4-x^2)-(1\over x+2)=0\)

ODZ: \(x-2≠0\)

\(4-x^2≠0\)

\(x+2≠0\)

It seems so far everything is simple. What's next? The next step will depend on how advanced you are in math. If you can, then solve these signed equationsand if you can't, leave it as it is for now. And we move on.

Further, all the fractions included in the equations must be represented as a single fraction. To do this, you need to find the common denominator of the fraction. And at the end write out what happened in the numerator and equate this expression to zero. And then solve the equation.

Let's go back to our examples:\((-2x-4\over x^2-4)=(x+5 \over x-2) \) ODZ: \(x^2-4≠0\)

\((-2x-4\over x^2-4)-(x+5 \over x-2)=0 \)\(x-2≠0 \)

Moved the fraction to the left, while changing the sign. We notice that the denominator\(x^2-4 \) can be factored using the reduced multiplication formula\(x^2-4=(x-2)(x+2)\) , and in the numerator you can take the common factor "-2" out of the bracket.

\((-2(x+2)\over (x+2)(x-2)) -(x+5\over x-2)=0\)

Once again we look at the ODZ, do we have it? There is! Then we can reduce the first fraction by x+2 . If there is no ODZ, it is impossible to reduce! We get:

\((-2\over x-2)-(x+5 \over x-2)=0\)

Fractions have a common denominator, so they can be subtracted:

\((-2-x-5\over x-2)=0\)

We pay attention, since we subtract the fractions, we change the “+” sign in the second fraction to a minus! We give in the numerator like terms:

\((-x-7 \over x-2)=0\)

Recall that a fraction is zero when the numerator is zero and the denominator is not zero. The fact that the denominator is not equal to zero, we indicated in the ODZ. It's time to indicate that the numerator is zero:

\(-x-7=0\)

This is a linear equation, move "-7" to the right, change the sign:

\(-x=7\)

\(x=7:(-1)\)

\(x=-7\)

Let's remember the ODZ:\(x^2-4≠0 \) \(x-2≠0\). If you were able to decide, then solve it like this:\(x^2≠4 \) \(x≠2\)

\(x_1≠2 \) \(x_2≠-2\)

And if they couldn’t decide, then we substitute in the ODZ instead of “x” what happened. We have\(x=-7\)

Then: \((-7)^2-4≠0\) ? Performed? Performed!

So the answer to our equation is:\(x=-7\)

Consider the following equation: \((3x^2-6\over 2(x+1))=(x-1)\)

We solve it the same way. First, we specify the ODZ:\(x+1≠0\)

Then transfer x-1 to the left, we immediately attribute the denominator 1 to this expression, this can be done, since the denominator 1 does not affect anything.

We get: \((3x^2-6\over 2(x+1)) -(x-1\over1)=0\)

Looking for a common denominator\(2(x+1)\) . We multiply the second fraction by this expression.

Got: \((3x^2-6\over2(x+1)) -((x-1)⋅2(x+1)\over2(x+1)) =0\)

\(( 3x^2-6-2x^2+2\over2(x+1)) =0 \)

If it's difficult, I'll explain:\(2(x+1)(x-1)=2x^2-2 \) And since there is a “-” sign before the second fraction, then, by combining these fractions into one, we change the signs to the opposite.

Notice that \(x^2-4=(x-2)(x+2)\) and rewrite it like this:\(((x-2)(x+2)\over2(x+1)) =0\)

Next, we use the definition of a fraction equal to zero. A fraction is zero when the numerator is zero and the denominator is not zero. The fact that the denominator is not equal to zero, we indicated in the ODZ, we indicate that the numerator is equal to zero.\((x-2)(x+2)=0\) . Let's solve this equation. It has two multipliers. x-2 and x+2 . Remember that the product of two factors is zero when one of the factors is zero.

So: x+2 =0 or x-2 =0

From the first equation we get x=-2 , from second x=2 . We transfer the number and change the sign.

At the last stage, we check the ODZ: x+1≠0

We substitute the numbers 2 and -2 for x.

We get 2+1≠0 . Performed? Yes! So x=2 is our root. We check the following:-2+1≠0 . Performed. Yes. Hence x=-2, also our root. So the answer is: 2 and -2.

We will solve the last equation without explanation. The algorithm is the same:

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Preview:

Lesson on the topic "Solution of fractional rational equations." 8th grade

Lesson Objectives:

Tutorial:

consolidation of the concept of a fractional rational equation;
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.

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 of critical thinking;
development of research skills.

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.

Lesson type : lesson - consolidation and systematization of knowledge, skills and abilities.

During the classes

1. Organizational moment.

Hello guys! Today in the lesson we will consider with you various ways to solve fractional rational equations. Equations are written on the blackboard, look at them carefully. Can you solve all of these equations?

1. 7 x - 14 = 0

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, solving equations

Please answer the following questions:

What is equation #1 called? ( Linear .) Method for solving 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).

Solve Equation #1

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.)

Solve Equation #3

What is Equation #2? ( Proportion ). 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.)

Solve Equation #2

Decision:

9 x \u003d 18 ∙ 5

9 x = 90

X = 90: 9

X = 10

Answer: 10

What fractional rational equation can you try to solve using the basic property of proportion? (No. 5). But since this equation has a denominator containing the unknown, it is necessary to write ...? ODZ.

Decision:

ODZ: x ≠ − 2, x ≠ 4

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

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

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

11 x = -2

X \u003d -2: (-11)

Answer:

Let's solve equation No. 4. What property is used to solve this equation? (If both sides of the equation are multiplied by the same non-zero number, then an equation is obtained that is equivalent to the given.)

Decision:

| ∙ 6

3 x - 3 + 4 x \u003d 5x

7 x - 5 x \u003d 3

2 x = 3

x=3:2

x = 1.5

Answer: 1.5

What fractional rational equation can be solved by multiplying both sides of the equation by the denominator? (No. 6).

Decision:

| ∙ (7 - x)

12 \u003d x (7 - x)

12 \u003d 7 x - x 2

x 2 - 7 x + 12 = 0

D \u003d 1\u003e 0, x 1 \u003d 3, x 2 \u003d 4.

Answer: 3; 4.

Now let's solve Equation #7 in two ways.

Decision:

1 way:

ODZ: x ≠ 0, x ≠ 5

When is a fraction equal to zero? (A fraction is zero when the numerator is zero and the denominator is non-zero.)

x ² - 3 x - 10 = 0

D \u003d 49\u003e 0, x 1 \u003d 5, x 2 \u003d - 2

X = 5 does not satisfy the ODZ. They say 5 is an extraneous root.

Answer: - 2

Decision:

2 way:

| ∙ x (x - 5) ODZ: x ≠ 0, x ≠ 5

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

x ² - 3 x + x - 5 - x - 5 \u003d 0

x ² - 3 x - 10 = 0

D \u003d 49\u003e 0, x 1 \u003d 5, x 2 \u003d - 2

X = 5 does not satisfy the ODZ. 5 - extraneous root.

Answer: - 2

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

Move everything to the left.
Bring fractions to a common denominator.
Solve the equation using the rule: a fraction is zero when the numerator is zero and the denominator is non-zero.
Exclude from its roots those that turn the denominator to zero (with the help of ODZ or by checking)
Write down the answer.

Another way to solve.

Algorithm for solving fractional rational equations:

1. Find the common denominator of the fractions included in the equation;

2. Multiply both sides of the equation by a common denominator; do not forget to write odz

3. Solve the resulting whole equation;

4. Eliminate from its roots those that turn the common denominator to zero (using DPV or by checking)

5. Write down the answer.

You can also solve the equation using the main property of the proportion, remembering to exclude from its roots those that turn the denominator to zero (using the ODZ or by checking)

8. Summing up the lesson.

So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations in various ways. In the next lesson, at home, you will have the opportunity to consolidate the acquired knowledge.

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.