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The rule of addition and subtraction of algebraic fractions. Addition of algebraic fractions

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Educational and training aids in the online store "Integral"
Manual for the textbook Muravina G.K. Manual for the textbook Makarychev Yu.N.

What is an algebraic fraction?

An algebraic fraction is an expression of the form: $\frac(P)(Q)$.

Where:
P is the numerator of an algebraic fraction.
Q is the denominator of an algebraic fraction.

Here are some examples algebraic fractions:

$\frac(a)(b)$, $\frac(12)(q-p)$, $\frac(7y-4)(y)$.

Basic properties of algebraic fractions

Property 1.
Both the numerator and denominator of a fraction can be multiplied by the same number (either by a monomial or by a polynomial). As a result, we will get the same fraction, but presented in a different form.

Otherwise, this transformation is called identical. It is used to bring an algebraic (and not only) expression to a simpler form, and working with this expression will be more convenient.

$\frac(a)(4b^2)=\frac(a*3b)(4b^2*3b)=\frac(3ab)(12b^3)$.

We have multiplied both the numerator and the denominator by the monomial $3b$. As a result, we got a fraction identical to the original one.

$\frac(a^2)(6b^3)=\frac(a^2*2)(6b^3*2)=\frac(2a^2)(12b^3)$.

If necessary, an algebraic fraction can be multiplied by a prime number. In this example, we multiplied both the numerator and denominator by the number 2. And again we got a fraction identical to the original one.

Property 2.
Both the numerator and the denominator of a fraction can be divided by the same number (either a monomial or a polynomial). As a result, we get the same fraction, but presented in a different form.

As in the case of multiplication, this identical transformation is resorted to to represent a fraction in more simple form and make it easier to work with.

Addition and subtraction of algebraic fractions with the same denominators

If algebraic fractions have the same denominators, they are added like ordinary fractions (only numerators are added, and the denominator remains common).

General rule:

$\frac(a)(d)+\frac(b)(d)-\frac(c)(d)=\frac(a+b-c)(d)$.

Example.

Simplify the expression:

$\frac(2a^2+5)(a^2-ab)+\frac(2ab+b)(a^2-ab)-\frac(b+5)(a^2-ab)$.

Solution.

We use the rule for adding fractions, which is described above, that is, we add the numerators, and write down the common denominator.

$\frac(2a^2+5)(a^2-ab)+\frac(2ab+b)(a^2-ab)-\frac(b+5)(a^2-ab)=\frac ((2a^2+5)+(2ab+b)-(b+5))(a^2-ab)$.

Let's work with the numerator.

$(2a^2+5)+(2ab+b)-(b+5)=$
$2a^2+5+2ab+b-b-5=2a^2+2ab$.

The result is a fraction:

$\frac(2a^2+2ab)(a^2-ab)$.

Guys, before finishing the solution, check if it is possible to further simplify the result. After all, this is the whole point of the transformation - to simplify the expression.
If you look closely, you can understand that the resulting fraction can be further simplified.

$\frac(2a^2+2ab)(a^2-ab)=\frac(2a(a+b))(a(a-b))=\frac(2(a+b))(a-b)=\ frac(2a+2b)(a-b)$.

Addition and subtraction of algebraic fractions with different denominators

When adding algebraic fractions with different denominators, you must act in the same way as when working with ordinary fractions. First you need to reduce the fraction to a common denominator, and then add or subtract the numerators of the fractions, in accordance with general rule which we have reviewed.

Example.
Calculate:

$\frac(a)(4b^2)+\frac(a^2)(6b^3)$.

Solution.
Let's bring these fractions to a common denominator. AT this example the common denominator is the monomial $12b^3$.
Then.

$\frac(a)(4b^2)+\frac(a^2)(6b^3)=\frac(3ab)(12b^3)+\frac(2a^2)(12b^3)=
\frac(3ab+2a^2)(12b^3)$.

The hardest part is finding the common denominator for the fractions. In some cases, this is not simple task.
When finding a common denominator, you can follow the rules:
1. If both denominators are monomials without brackets, then it is better to choose a common denominator for the number first, and then for the variable. In our example, the number is 12 and the variable is $b^3$.
2. If the denominator is a more complex expression, for example, $x + 1$, $x + y$ and the like, then it is better to choose the denominator in the form of a product of denominators, for example, $ (x + y) (x - y) $. Such a denominator is divisible by both $x + y$ and $x - y$.

Remember!
For two algebraic fractions of common denominators, you can choose as many as you like. But to simplify the calculations, you need to choose the simplest possible.

form the ability to perform actions (addition and subtraction) with algebraic fractions with different denominators, based on the rule of addition and subtraction ordinary fractions with different denominators;

repeat and consolidate the addition and subtraction of fractions with the same denominators.

Equipment: Demonstration material.

Tasks for updating knowledge:

1) +; 2) -;

3) + ; 4) +; 5) -.

1) Algorithm for adding and subtracting ordinary fractions with different denominators.

To add or subtract common fractions with different denominators:

Convert these fractions to the smallest common denominator.
Add or subtract the resulting fractions.

2) Algorithm for reducing algebraic fractions to a common denominator.

Let's find additional factors for each of the fractions: these will be the products of those factors that are in the common (new) denominator, but which are not in the old denominator.

3) Standards for independent work with self-test:

3) Card for the reflection stage.

This topic is clear to me.
I know how to find additional factors for each of the fractions.
I can find new numerators for each of the fractions.
In independent work, I succeeded.
I was able to understand the reason for the mistake that I made in my independent work.
I am satisfied with my work in the classroom.

DURING THE CLASSES

1. Self-determination to activity.

Stage goals:

Inclusion of students in learning activities: continuation of the journey through the country "Algebraic Expressions".
Determining the content of the lesson: continuing to work with algebraic fractions.

Organization educational process at step 1:

Good morning, guys! We continue our fascinating journey through the country "Algebraic Expressions".

What “inhabitants” of the country did we meet in previous lessons? (With algebraic expressions.)

What can we do with familiar algebraic expressions? (Addition and subtraction.)

Which salient feature algebraic fractions that we already know how to add and subtract? (We add and subtract fractions that have the same denominator.)

Right. But we all understand well together that the skills to perform actions with algebraic fractions that have the same denominators are not enough. What else do you think we need to learn to do? (Perform actions with fractions that have different denominators.)

Well done! Shall we continue our journey then? (Yes!)

2. Actualization of knowledge and fixation of difficulties in activity.

Stage goals:

Update knowledge about performing actions with fractions with the same denominators, methods of oral calculations.
Fix difficulty.

Organization of the educational process at stage 2:

There are several examples on the board for performing actions with fractions:

5) -=-==.

Students are encouraged to voice their solutions in a loud speech.

In the first example, the guys easily give the correct answer, remembering the algorithm for performing actions with algebraic fractions that have the same denominators.

When the comment on example #2 has already been made, the teacher focuses on example #2:

Guys, look what we have interesting in example number 2? (We not only performed actions with algebraic fractions that have the same denominators, but also performed the reduction of the resulting algebraic fraction: we took the minus sign out of brackets, we got the same factors in the numerator and denominator, by which we subsequently reduced the result.)

It is very good that you have not forgotten that the basic property of a fraction is applicable not only to ordinary, but also to algebraic fractions!

Who will comment on the solution of the following three examples for everyone?

Most likely, there will be a student who can easily solve example number 3.

What did you use when solving example number 3? (The algorithm for adding and subtracting ordinary fractions with different denominators helped me.)

How exactly did you act? (I reduced algebraic fractions to the lowest common denominator of 15 and then added them.)

Wonderful! And how are we doing with the last two examples?

When it comes to the next two examples, the guys (each for themselves) fix the difficulty that has arisen.

The words of the students are something like this:

I find it difficult to complete examples 4-5, because before me are algebraic fractions, not with “same” denominators, and these different denominators include variables (No. 4), and in No. 5 there are literal expressions in the denominators in general! ..”

Answers to tasks 4–5 were not received.

3. Identification of the place and causes of difficulties and setting the goal of the activity.

Stage goals:

Fix distinguishing feature tasks that caused difficulty in learning activities.
State the purpose and topic of the lesson.

Organization of the educational process at stage 3:

Guys? Where did the difficulty arise? (In examples 4-5.)

Why, when solving them, are you not ready to discuss the solution and give an answer? (Because the algebraic fractions proposed in these tasks have different denominators, and we are familiar with the algorithm for performing operations with algebraic fractions that have the same denominators.

What else do we need to be able to do? (You need to learn how to add and subtract fractions with different denominators.)

I agree with you. How can we formulate the topic of our today's lesson? (Addition and subtraction of algebraic fractions with different denominators.)

The topic of the lesson is written in notebooks.

4. Building a project to get out of the difficulty.

Purpose of the stage:

Children building a new way of doing things.
Fixing the algorithm for reducing algebraic fractions to a common denominator.

Organization of the educational process at stage 4:

What is the purpose of our lesson today? (Learn to add and subtract algebraic fractions with different denominators.)

How to be? (For this we have to build an algorithm further work with algebraic fractions.)

What do we need to come up with to achieve the goal of the lesson? (An algorithm for reducing algebraic fractions to a common denominator, so that later we can work according to the usual rule for adding and subtracting fractions with the same denominators.)

The work can be organized in groups, each group is given a piece of paper and a marker. Students can offer their own variants of the algorithm in the form of a list of steps. You have 5 minutes to work. Groups post their options for an algorithm or rule, and then each option is analyzed.

Most likely, one of the students will definitely draw an analogy of their algorithm with the algorithm for adding and subtracting ordinary fractions with different denominators: first, they bring the fractions to a common denominator using the appropriate additional factors, and then add and subtract the resulting fractions with the same denominators.

Subsequently, a single variant is derived from this. It can be like this:

We decompose all the denominators into factors.
From the first denominator we write out the product of all its factors, from the remaining denominators we assign the missing factors to this product. The resulting product will be the common (new) denominator.
Let's find additional factors for each of the fractions: these will be the products of those factors that are in the new denominator, but which are not in the old denominator.
Let's find a new numerator for each fraction: it will be the product of the old numerator and an additional factor.
Let's write each fraction with a new numerator and a common (new) denominator.

Well, let's apply our rule to complete the unsolved proposed tasks. Each task (4, 5) is spoken in turn by some students of the class, the teacher fixes the solution on the board.

We are simply geniuses! We have built an algorithm for adding and subtracting algebraic fractions with different denominators. By joint efforts, we have eliminated the difficulty, since we now have a real “guide” (algorithm) in the country unknown to us “Algebraic fractions”!

5. Primary consolidation in external speech.

Purpose of the stage:

Train the ability to bring algebraic fractions to a common denominator.
Organize the pronunciation of the studied content of the rule-algorithm in external speech.

Organization of the educational process at stage 5:

Guys, but we all know well that just looking and knowing the “map of the area” is not a journey. What should we do to penetrate deeper and more into the world of algebraic fractions? (We have to solve examples, and generally practice solving examples, in order to consolidate our new algorithm.)

Quite right. Therefore, I propose to begin our study.

The student verbally pronounces the plan of his decision, the teacher corrects if some inaccuracies are made.

Approximately it sounds like this:

We must choose a number that will be divided by 2 and 5 at the same time. This is the number 10. Then we select the variables to the degree we need. So our new denominator will be 10xy. We select additional multipliers. To the first fraction: 5y, to the second: 2x. We multiply the selected additional factors by each old numerator. We get algebraic fractions with the same denominators, perform the subtraction according to the rule already familiar to us.

I'm happy. And now our big team will split into pairs, and we will continue our interesting path.

No. 133 (a, d). The students work in pairs, saying the solution to each other:

a) +=+= =;

d) +=+= =.

6. Independent work with self-test.

Stage goals:

Spend independent work.
Perform a self-test against the prepared self-test standard.
Students will record difficulties, identify the causes of errors and correct errors.

Organization of the educational process at stage 6:

I carefully observed your work and came to the conclusion that each of you is already ready to independently think about ways and find solutions to examples on our today's topic. Therefore, I offer you a small independent work, after which you will be offered a standard with the correct solution and answer.

No. 134 (a, b): perform work on options.

After the work is completed, a standard check is carried out. When checking solutions, students mark “+” the correct solution, “?” not the right decision. It is desirable that students who make mistakes explain the reason why they did the task incorrectly.

Errors are analyzed and corrected.

So, what difficulties did you meet on your way? (I made a mistake when opening brackets that are preceded by a minus sign.)

What is the reason for this? (Simply due to inattention, but in the future I will be more careful!)

What else seemed difficult? (Did I have a hard time finding additional factors for fractions?)

You should definitely study step 3 of the algorithm in more detail so that such a problem does not arise in the future!

Were there any other difficulties? (And I just did not bring similar terms).

And we'll fix it. When you have done everything that is possible according to the new algorithm, you need to remember the material studied for a long time. In particular, the reduction of similar terms, or the reduction of fractions, etc.

7. Inclusion of new knowledge in the knowledge system.

The purpose of the stage: to repeat and consolidate the algorithm for adding and subtracting algebraic fractions with different denominators studied in the lesson.

8. Lesson reflection.

The purpose of the stage: to fix the new content, evaluate their own activities.

Organization of the educational process at stage 8:

What was our goal at the beginning of the lesson? (Learn how to add and subtract fractions with different denominators.)

What did we come up with to achieve the goal? (An algorithm for adding and subtracting algebraic fractions with different denominators.)

What else did we use? (We factored the denominators, selected LCMs for the coefficients, and additional factors for the numerators.)

Now take some colored pen or felt-tip pen and mark with a “+” sign those statements with the truth of which you agree:

Each student has a card with phrases. Children mark and show to the teacher.

Well done!

Homework: paragraph 4 (textbook); No. 126, 127 (task book).

The video lesson "Addition and subtraction of algebraic fractions with different denominators" is visual aid, with the help of which theoretical material is given, the algorithms and features of performing operations of subtraction, addition of fractions with different denominators are explained in detail. With the help of the manual, it is easier for the teacher to form the ability of students to perform operations with algebraic fractions. During the video tutorial, a number of examples are considered, the solution of which is described in detail, paying attention to important details.

The use of a video lesson in a mathematics lesson enables the teacher to achieve learning goals faster and increase the effectiveness of learning. The visibility of the demonstration helps students to remember the material, to master it more deeply, so the video can be used to accompany the teacher's explanation. If this video is used as part of the lesson, then the teacher's time is freed up for strengthening individual work and the use of other learning tools to improve learning efficiency.

The demo starts by introducing the topic of the video tutorial. It is noted that the performance of operations of subtraction, addition of algebraic fractions is similar to the performance of operations with ordinary fractions. The mechanism of subtraction, addition for ordinary fractions is recalled - fractions are reduced to a common denominator, after which the operations themselves are performed directly.

The algorithm of subtraction, addition of algebraic fractions is voiced and described on the screen. It consists of two steps - reducing fractions to the same denominators and then performing the addition (or subtraction) of fractions with equal denominators. The application of the algorithm is considered on the example of finding the values of the expressions a/4b 2 -a 2 /6b 3 , as well as x/(x+y)-x/(x-y). It is noted that to solve the first example, it is necessary to bring both fractions to the same denominator. This denominator will be 12b 3 . Bringing these fractions to the denominator 12b 3 was discussed in detail in the last video tutorial. The transformation results in two fractions with equal denominators 3ab/12b 3 and 2a 2 /12b 3 . These fractions are added according to the rule for adding fractions with equal denominators. After adding the numerators of the fractions, the result is the fraction (3ab+2a 2)/12b 3 . The following describes the solution of the example x/(x+y)-x/(x-y). After reducing the fractions to the same denominator, the fractions (x 2 -xy) / (x 2 -y 2) and (x 2 + xy) / (x 2 -y 2) are obtained. According to the rule for subtracting fractions with equal denominators, we perform an operation with numerators, after which we get a fraction -2xy / (x 2 -y 2).

It is noted that the most difficult step in solving problems of addition, subtraction of fractions with different denominators is their reduction to a common denominator. Tips are given on how to easily develop skills in solving these problems. Understand the common denominator of a fraction. It consists of a numerical coefficient with a variable raised to a power. It can be seen that the expression can be divided by the denominators of the first and second fractions. In this case, the numerical coefficient 12 is the least common multiple of the numerical coefficients of fractions 4 and 6. And the variable b contains both denominators 4b 2 and 6b 3 . In this case, the common denominator contains the variable to the greatest extent among the denominators of the original fractions. Also considered is finding a common denominator for x/(x+y) and x/(x-y). It is noted that the common denominator (x+y)(x-y) is divided by each denominator. So, the solution of the problem comes down to finding the least common multiple of the available numerical coefficients, as well as finding the highest exponent for a letter variable that occurs several times. Then, after collecting these parts into a common product, a common denominator is obtained.

An algorithm for finding a common denominator for several fractions is voiced and formulated on the screen. This algorithm consists of four stages, in the first of which the denominators are factorized. At the second stage of the algorithm, the least common multiple of the available data of the coefficients included in the denominators of the fractions is found. At the third stage, a product is compiled, which includes the literal factors of the expansions of the denominators, while the literal indicator present in several denominators is chosen to the greatest extent. At the fourth stage, the numerical and alphabetic factors found in the previous stages are collected into one product. This will be the common denominator. A remark is made to the considered algorithm. In the example of finding the common denominator of fractions a / 4b 2 and a 2 /6b 3, it is noted that in addition to 12b 3 there are other denominators 24b 3 and 48a 2 b 3 . And for every set of fractions, there are many common denominators. However, the denominator 12b 3 is the simplest and most convenient, so it is also called the least common denominator of the original fractions. Additional factors are the result of the partial common denominator and the original denominator of the fraction. Demonstrated in detail through animation, how the numerator, denominator of fractions is multiplied by an additional factor.

Further, it is proposed to consider the algorithm for reducing algebraic fractions to a common denominator in a simpler form, so that it is more understandable for students. It also consists of four steps, the first of which is the factorization of the denominators. Then it is proposed to write out all the factors from the first denominator, to supplement the product with the missing factors from the remaining denominators. Thus, a common denominator is found. Additional factors are found for each fraction from those factors of the denominator that did not fall into the common denominator. The fourth step is to determine for each fraction a new numerator, which is the product of the old numerator and an additional factor. Then each fraction is written with a new numerator and denominator.

The following example describes a simplification of the expression 3a/(4a 2 -1)-(a+1)/(2a 2 +a). At the first stage of the solution, the denominators of each fraction are decomposed into factors. For products, the common factor is (2a + 1). Complementing the product with the remaining factors (2a-1) and a, a common denominator of the form a (2a-1) (2a + 1) is obtained. under construction auxiliary table, which indicates the common denominator, denominators, additional factors. At the second stage of the solution, each numerator is multiplied by an additional factor, subtraction is performed. The result is a fraction (a 2 -a + 1) / a (2a-1) (2a + 1).

Example 3 considers a simplification of the expression b/(2a 4 +4a 3 b+2a 2 b 2)-1/(3ab 2 -3a 3)+b/(6a 4 -6a 3 b). The solution is also analyzed in stages, attention is drawn to the essential features of the operations, the reduction of fractions to a common denominator, the performance of operations with the numerator are described in detail. As a result of calculations and after transformation, a fraction is obtained (2a 3 +6a 2 b-ab 2 +b 3)/6a 3 (a-b)(a+b) 2 .

The video lesson "Addition and subtraction of algebraic fractions with different denominators" can serve as a means of increasing the effectiveness of a mathematics lesson on this topic. The manual will be useful to the teacher who distance learning, for visualization educational material. For students, a video lesson can be recommended for self-study, as it explains in detail and clearly the features of performing the operations being studied.

Lesson topic: Addition and subtraction of algebraic fractions.

Lesson Objectives:

Tutorials:

review the rules for adding and subtracting fractions with the same denominator
introduce rules for adding and subtracting algebraic fractions with the same denominators;
to form the ability to perform addition and subtraction with algebraic fractions.

Developing:

develop thinking, attention, memory, the ability to analyze, compare, compare;
expanding the horizons of students;

vocabulary replenishment;

Educational:

bring up cognitive interest to the subject.
Cultivate a culture of intellectual work

Equipment:

cards - test tasks;
a computer;
projector;
screen;
lesson presentation

Motto:

You can't learn math by watching your neighbor do it!

Slide 2.

Lesson plan.

Reporting the purpose and topic of the lesson (2 min);
Update basic knowledge and skills of students (4 min);
Oral work (5 min);
Learning new material (8 min);
Physical education (2 min);
Consolidation of new material (10 min);
Multiple choice test (10 min);
The result of the lesson, conclusions (2 min);
Homework. (2 minutes).

Slide 3.

During the classes.

I. Organizational moment:

1) message of the topic of the lesson;

2) communication of the goals and objectives of the lesson.

II. Knowledge update:

What is an algebraic fraction? Give examples.

What does it mean to reduce an algebraic fraction?

How to bring algebraic fractions to a common denominator?

slide 4.

III. Oral work:

Read fractions:
Find an expression that is redundant a) (a + c) 2; b) ; in) ; G) .
Restore partially erased records: to reduce to a common denominator

Slide 5.

find the mistake

slide 6.

For each fraction, find the fraction equal to it, using the correspondence number - letter:

1) ; 2) 3) .

A) b); in) .

slide 7.8

IV. Learning new material.
1) Repeat the rules for adding and subtracting numerical fractions with the same denominators. Then verbally solve the following examples:

2) Remember the rules for adding and subtracting polynomials and write the following exercises on the board:

3) Students should suggest rules for doing the following examples written on the board:

The solution of the examples is discussed. If the students cannot cope on their own, the teacher explains.

slide 9.

The rules for adding and subtracting algebraic fractions with the same denominators are written in a notebook.
, .

slide 10.

V. Physical education for the eyes

Exercise 1. Make 15 oscillatory movements of the eyes horizontally from right to left, then from left to right.

Exercise 2. Make 15 oscillatory eye movements vertically up - down and down - up.

Exercise 3. Also 15, but circular rotational movements eyes from left to right.

Exercise 4. The same, but from right to left.

Exercise 5. Make 15 circular rotational movements with your eyes, first to the right, then to left side, as if drawing a figure eight laid on its side with its eyes.

VI. Consolidation of new material.
1) Front work.

1) Solve tasks

№ 462 (1,3)

2) Add fractions:

3) Subtract fractions:

4) Perform actions.

Slide 11.

2) Individual work.
Four students perform independent work on the board, proposed on the cards.

Card 1.

Card 2.

Card 3.

Card 4.

The rest in notebooks: Perform addition and subtraction of fractions:
a) b)
in)

VII. Performing work in groups and analyzing the results.

Each group is given test tasks, after completing which they receive a word - the name of a famous mathematician.

Exercise	Possible answer	Letter







	x + 10

Exercise	Possible answer	Letter

Exercise	Possible answer	Letter

Exercise	Possible answer	Letter

Answer table:

job number
Letter

Check the quality of the job.

Did you get the name of a famous mathematician from the received letters?

If you answered all the questions correctly, you got an “EXCELLENT” rating!!!

If you made a mistake in one step - not bad, but the scientist would probably be offended. You have been rated "GOOD"!

If you made a mistake in two steps, then you did not listen well to the teacher in the lesson and you will have to read the topic in the algebra textbook. You have been rated "SATISFACTORY".

If you made a mistake in more than two steps, then you did not listen to the teacher at all in the lesson and you will have to read the algebra textbook very carefully. You have been rated "UNSATISFACTORY".

Slide 13-17.

When time is available, tasks are solved:
1. Prove that the expression
for all values of a2 takes positive values.
2. Present a fraction as a sum or difference of an integer expression and a fraction:
a); b) c)

3. Knowing that, find the value of the fraction:
a); b) c)

VIII. Summarizing.

I X. Homework:Read the textbook material p.26, learn the rules of this paragraph. Solve problems No. 462(2,4); make 5 examples for adding and subtracting algebraic fractions; find information about the mathematicians whose names we heard today.

How to perform addition of algebraic (rational) fractions?

To add algebraic fractions, you need:

1) Find the smallest of these fractions.

2) Find an additional factor for each fraction (for this you need to divide the new denominator by the old one).

3) Multiply the additional factor by the numerator and denominator.

4) Perform the addition of fractions with the same denominators

(To add fractions with the same denominators, you must add their numerators, and leave the denominator the same).

Examples of addition of algebraic fractions.

The lowest common denominator is the sum of all factors taken to the highest power. AT this case it is equal to ab.

To find an additional factor to each fraction, we divide the new denominator by the old one. ab:a=b, ab:(ab)=1.

The numerator has a common factor a. We take it out of the bracket and reduce the fraction by a:

The denominators of these fractions are polynomials, so they need to be tried. In the denominator of the first fraction there is a common factor x, in the second - 5. We take them out of brackets:

The common denominator consists of all factors included in the denominator and is equal to 5x(x-5).

To find an additional factor to each fraction, we divide the new denominator by the old one.

(If you don’t like the division, you can do it differently. We reason like this: what do you need to multiply the old denominator to get a new one? To get 5x(x-5) from x (x-5), you need to multiply the first expression by 5. To get from 5 (x-5) to get 5x(x-5), you need to multiply the 1st expression by x. Thus, the additional factor to the first fraction is 5, to the second - x).

The numerator is the full square of the difference. We collapse it according to the formula and reduce the fraction by (x-5):

The denominator of the first fraction is a polynomial. It does not factor into factors, so the common denominator of these fractions is equal to the product of the denominators m (m + 3):

Polynomials in the denominators of fractions,. We take out the common factor x in the denominator of the first fraction, and 2 in the denominator of the second fraction: