Complex examples with ordinary fractions. Fractions, operations with fractions

This article examines operations on fractions. Rules for addition, subtraction, multiplication, division or exponentiation of fractions of the form A B will be formed and justified, where A and B can be numbers, numerical expressions or expressions with variables. In conclusion, examples of solutions with detailed descriptions will be considered.

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Rules for performing operations with general numerical fractions

Numerical fractions general view have a numerator and denominator in which there are integers or numeric expressions. If we consider fractions such as 3 5, 2, 8 4, 1 + 2 3 4 (5 - 2), 3 4 + 7 8 2, 3 - 0, 8, 1 2 2, π 1 - 2 3 + π, 2 0, 5 ln 3, then it is clear that the numerator and denominator can have not only numbers, but also expressions of various types.

Definition 1

There are rules by which actions are performed with ordinary fractions. It is also suitable for general fractions:

• When subtracting fractions with like denominators, only the numerators are added, and the denominator remains the same, namely: a d ± c d = a ± c d, the values ​​a, c and d ≠ 0 are some numbers or numerical expressions.
• When adding or subtracting a fraction with different denominators, it is necessary to reduce it to a common denominator, and then add or subtract the resulting fractions with the same exponents. Literally it looks like this: a b ± c d = a · p ± c · r s, where the values ​​a, b ≠ 0, c, d ≠ 0, p ≠ 0, r ≠ 0, s ≠ 0 are real numbers, and b · p = d · r = s . When p = d and r = b, then a b ± c d = a · d ± c · d b · d.
• When multiplying fractions, the action is performed with numerators, after which with denominators, then we get a b · c d = a · c b · d, where a, b ≠ 0, c, d ≠ 0 act as real numbers.
• When dividing a fraction by a fraction, we multiply the first by the second inverse, that is, we swap the numerator and denominator: a b: c d = a b · d c.

Rationale for the rules

There are the following mathematical points that you should rely on when calculating:

• the slash means the division sign;
• division by a number is treated as multiplication by its reciprocal value;
• application of the property of operations with real numbers;
• application of the basic property of fractions and numerical inequalities.

With their help, you can perform transformations of the form:

a d ± c d = a · d - 1 ± c · d - 1 = a ± c · d - 1 = a ± c d ; a b ± c d = a · p b · p ± c · r d · r = a · p s ± c · e s = a · p ± c · r s ; a b · c d = a · d b · d · b · c b · d = a · d · a · d - 1 · b · c · b · d - 1 = = a · d · b · c · b · d - 1 · b · d - 1 = a · d · b · c b · d · b · d - 1 = = (a · c) · (b · d) - 1 = a · c b · d

Examples

In the previous paragraph it was said about operations with fractions. It is after this that the fraction needs to be simplified. This topic was discussed in detail in the paragraph on converting fractions.

First, let's look at an example of adding and subtracting fractions with the same denominator.

Example 1

Given the fractions 8 2, 7 and 1 2, 7, then according to the rule it is necessary to add the numerator and rewrite the denominator.

Solution

Then we get a fraction of the form 8 + 1 2, 7. After performing the addition, we obtain a fraction of the form 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3. So, 8 2, 7 + 1 2, 7 = 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3.

Answer: 8 2 , 7 + 1 2 , 7 = 3 1 3

There is another solution. To begin with, we switch to the form of an ordinary fraction, after which we perform a simplification. It looks like this:

8 2 , 7 + 1 2 , 7 = 80 27 + 10 27 = 90 27 = 3 1 3

Example 2

Let's subtract from 1 - 2 3 · log 2 3 · log 2 5 + 1 a fraction of the form 2 3 3 · log 2 3 · log 2 5 + 1 .

Since equal denominators are given, it means that we are calculating a fraction with the same denominator. We get that

1 - 2 3 log 2 3 log 2 5 + 1 - 2 3 3 log 2 3 log 2 5 + 1 = 1 - 2 - 2 3 3 log 2 3 log 2 5 + 1

There are examples of calculating fractions with different denominators. An important point is reduction to a common denominator. Without this, we will not be able to perform further operations with fractions.

The process is vaguely reminiscent of reduction to a common denominator. That is, the least common divisor in the denominator is searched for, after which the missing factors are added to the fractions.

If the fractions being added do not have common factors, then their product can become one.

Example 3

Let's look at the example of adding fractions 2 3 5 + 1 and 1 2.

Solution

In this case, the common denominator is the product of the denominators. Then we get that 2 · 3 5 + 1. Then, when setting additional factors, we have that for the first fraction it is equal to 2, and for the second it is 3 5 + 1. After multiplication, the fractions are reduced to the form 4 2 · 3 5 + 1. The general reduction of 1 2 will be 3 5 + 1 2 · 3 5 + 1. We add the resulting fractional expressions and get that

2 3 5 + 1 + 1 2 = 2 2 2 3 5 + 1 + 1 3 5 + 1 2 3 5 + 1 = = 4 2 3 5 + 1 + 3 5 + 1 2 3 5 + 1 = 4 + 3 5 + 1 2 3 5 + 1 = 5 + 3 5 2 3 5 + 1

Answer: 2 3 5 + 1 + 1 2 = 5 + 3 5 2 3 5 + 1

When we are dealing with general fractions, then we usually do not talk about the lowest common denominator. It is unprofitable to take the product of the numerators as the denominator. First you need to check if there is a number that is less in value than their product.

Example 4

Let's consider the example of 1 6 · 2 1 5 and 1 4 · 2 3 5, when their product is equal to 6 · 2 1 5 · 4 · 2 3 5 = 24 · 2 4 5. Then we take 12 · 2 3 5 as the common denominator.

Let's look at examples of multiplying general fractions.

Example 5

To do this, you need to multiply 2 + 1 6 and 2 · 5 3 · 2 + 1.

Solution

Following the rule, it is necessary to rewrite and write the product of the numerators as a denominator. We get that 2 + 1 6 2 5 3 2 + 1 2 + 1 2 5 6 3 2 + 1. Once a fraction has been multiplied, you can make reductions to simplify it. Then 5 · 3 3 2 + 1: 10 9 3 = 5 · 3 3 2 + 1 · 9 3 10.

Using the rule for transition from division to multiplication by a reciprocal fraction, we obtain a fraction that is the reciprocal of the given one. To do this, the numerator and denominator are swapped. Let's look at an example:

5 3 3 2 + 1: 10 9 3 = 5 3 3 2 + 1 9 3 10

Then they must multiply and simplify the resulting fraction. If necessary, get rid of irrationality in the denominator. We get that

5 3 3 2 + 1: 10 9 3 = 5 3 3 9 3 10 2 + 1 = 5 2 10 2 + 1 = 3 2 2 + 1 = 3 2 - 1 2 2 + 1 2 - 1 = 3 2 - 1 2 2 2 - 1 2 = 3 2 - 1 2

Answer: 5 3 3 2 + 1: 10 9 3 = 3 2 - 1 2

This paragraph is applicable when a number or numerical expression can be represented as a fraction with a denominator equal to 1, then the operation with such a fraction is considered a separate paragraph. For example, the expression 1 6 · 7 4 - 1 · 3 shows that the root of 3 can be replaced by another 3 1 expression. Then this entry will look like multiplying two fractions of the form 1 6 · 7 4 - 1 · 3 = 1 6 · 7 4 - 1 · 3 1.

Performing Operations on Fractions Containing Variables

The rules discussed in the first article are applicable to operations with fractions containing variables. Consider the subtraction rule when the denominators are the same.

It is necessary to prove that A, C and D (D not equal to zero) can be any expressions, and the equality A D ± C D = A ± C D is equivalent to its range of permissible values.

It is necessary to take a set of ODZ variables. Then A, C, D must take the corresponding values ​​a 0 , c 0 and d 0. Substitution of the form A D ± C D results in a difference of the form a 0 d 0 ± c 0 d 0 , where, using the addition rule, we obtain a formula of the form a 0 ± c 0 d 0 . If we substitute the expression A ± C D, then we get the same fraction of the form a 0 ± c 0 d 0. From here we conclude that the selected value that satisfies the ODZ, A ± C D and A D ± C D are considered equal.

For any value of the variables, these expressions will be equal, that is, they are called identically equal. This means that this expression is considered a provable equality of the form A D ± C D = A ± C D .

Examples of adding and subtracting fractions with variables

When you have the same denominators, you only need to add or subtract the numerators. This fraction can be simplified. Sometimes you have to work with fractions that are identically equal, but at first glance this is not noticeable, since some transformations must be performed. For example, x 2 3 x 1 3 + 1 and x 1 3 + 1 2 or 1 2 sin 2 α and sin a cos a. Most often, a simplification of the original expression is required in order to see the same denominators.

Example 6

Calculate: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) , x - 1 x - 1 + x x + 1 .

Solution

1. To make the calculation, you need to subtract fractions that have the same denominator. Then we get that x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + 1 - 5 - x x + x - 2 . After which you can expand the brackets and add similar terms. We get that x 2 + 1 - 5 - x x + x - 2 = x 2 + 1 - 5 + x x + x - 2 = x 2 + x - 4 x + x - 2
2. Since the denominators are the same, all that remains is to add the numerators, leaving the denominator: l g 2 x + 4 x (l g x + 2) + 4 l g x x (l g x + 2) = l g 2 x + 4 + 4 x (l g x + 2)
   The addition has been completed. It can be seen that it is possible to reduce the fraction. Its numerator can be folded using the formula for the square of the sum, then we get (l g x + 2) 2 from abbreviated multiplication formulas. Then we get that
   l g 2 x + 4 + 2 l g x x (l g x + 2) = (l g x + 2) 2 x (l g x + 2) = l g x + 2 x
3. Given fractions of the form x - 1 x - 1 + x x + 1 with different denominators. After the transformation, you can move on to addition.

Let's consider a twofold solution.

The first method is that the denominator of the first fraction is factorized using squares, with its subsequent reduction. We get a fraction of the form

x - 1 x - 1 = x - 1 (x - 1) x + 1 = 1 x + 1

So x - 1 x - 1 + x x + 1 = 1 x + 1 + x x + 1 = 1 + x x + 1 .

In this case, it is necessary to get rid of irrationality in the denominator.

1 + x x + 1 = 1 + x x - 1 x + 1 x - 1 = x - 1 + x x - x x - 1

The second method is to multiply the numerator and denominator of the second fraction by the expression x - 1. Thus, we get rid of irrationality and move on to adding fractions with the same denominator. Then

x - 1 x - 1 + x x + 1 = x - 1 x - 1 + x x - 1 x + 1 x - 1 = = x - 1 x - 1 + x x - x x - 1 = x - 1 + x · x - x x - 1

Answer: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + x - 4 x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) = l g x + 2 x, 3) x - 1 x - 1 + x x + 1 = x - 1 + x · x - x x - 1 .

In the last example we found that reduction to a common denominator is inevitable. To do this, you need to simplify the fractions. When adding or subtracting, you always need to look for a common denominator, which looks like the product of the denominators with additional factors added to the numerators.

Example 7

Calculate the values ​​of the fractions: 1) x 3 + 1 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) (2 x - 4) - sin x x 5 ln (x + 1) (2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x

Solution

1. The denominator does not require any complex calculations, so you need to choose their product of the form 3 x 7 + 2 · 2, then choose x 7 + 2 · 2 for the first fraction as an additional factor, and 3 for the second. When multiplying, we get a fraction of the form x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 3 x 7 + 2 2 + 3 1 3 x 7 + 2 2 = = x x 7 + 2 2 + 3 3 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2
2. It can be seen that the denominators are presented in the form of a product, which means that additional transformations are unnecessary. The common denominator will be considered to be a product of the form x 5 · ln 2 x + 1 · 2 x - 4 . Hence x 4 is an additional factor to the first fraction, and ln(x + 1) to the second. Then we subtract and get:
   x + 1 x · ln 2 (x + 1) · 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x + 1 · x 4 x 5 · ln 2 (x + 1 ) · 2 x - 4 - sin x · ln x + 1 x 5 · ln 2 (x + 1) · (2 ​​x - 4) = = x + 1 · x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4) = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4 )
3. This example makes sense when working with fraction denominators. It is necessary to apply the formulas for the difference of squares and the square of the sum, since they will make it possible to move on to an expression of the form 1 cos x - x · cos x + x + 1 (cos x + x) 2. It can be seen that the fractions are reduced to a common denominator. We get that cos x - x · cos x + x 2 .

Then we get that

1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x = = 1 cos x - x cos x + x + 1 cos x + x 2 = = cos x + x cos x - x cos x + x 2 + cos x - x cos x - x cos x + x 2 = = cos x + x + cos x - x cos x - x cos x + x 2 = 2 cos x cos x - x cos x + x 2

Answer:

1) x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · ( 2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 · cos x · x + x = 2 · cos x cos x - x · cos x + x 2 .

Examples of multiplying fractions with variables

When multiplying fractions, the numerator is multiplied by the numerator and the denominator by the denominator. Then you can apply the reduction property.

Example 8

Multiply the fractions x + 2 · x x 2 · ln x 2 · ln x + 1 and 3 · x 2 1 3 · x + 1 - 2 sin 2 · x - x.

Solution

Multiplication needs to be done. We get that

x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = = x - 2 x 3 x 2 1 3 x + 1 - 2 x 2 ln x 2 ln x + 1 sin (2 x - x)

The number 3 is moved to the first place for the convenience of calculations, and you can reduce the fraction by x 2, then we get an expression of the form

3 x - 2 x x 1 3 x + 1 - 2 ln x 2 ln x + 1 sin (2 x - x)

Answer: x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = 3 x - 2 x x 1 3 x + 1 - 2 ln x 2 · ln x + 1 · sin (2 · x - x) .

Division

Division of fractions is similar to multiplication, since the first fraction is multiplied by the second reciprocal. If we take for example the fraction x + 2 x x 2 ln x 2 ln x + 1 and divide by 3 x 2 1 3 x + 1 - 2 sin 2 x - x, then it can be written as

x + 2 · x x 2 · ln x 2 · ln x + 1: 3 · x 2 1 3 · x + 1 - 2 sin (2 · x - x) , then replace with a product of the form x + 2 · x x 2 · ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x)

Exponentiation

Let's move on to considering operations with general fractions with exponentiation. If there is a power with a natural exponent, then the action is considered as multiplication of equal fractions. But it is recommended to use a general approach based on the properties of degrees. Any expressions A and C, where C is not identically equal to zero, and any real r on the ODZ for an expression of the form A C r the equality A C r = A r C r is valid. The result is a fraction raised to a power. For example, consider:

x 0, 7 - π · ln 3 x - 2 - 5 x + 1 2, 5 = = x 0, 7 - π · ln 3 x - 2 - 5 2, 5 x + 1 2, 5

Procedure for performing operations with fractions

Operations on fractions are performed according to certain rules. In practice, we notice that an expression may contain several fractions or fractional expressions. Then it is necessary to perform all actions in strict order: raise to a power, multiply, divide, then add and subtract. If there are parentheses, the first action is performed in them.

Example 9

Calculate 1 - x cos x - 1 c o s x · 1 + 1 x .

Solution

Since we have the same denominator, then 1 - x cos x and 1 c o s x, but subtractions cannot be performed according to the rule; first, the actions in parentheses are performed, then multiplication, and then addition. Then when calculating we get that

1 + 1 x = 1 1 + 1 x = x x + 1 x = x + 1 x

When substituting the expression into the original one, we get that 1 - x cos x - 1 cos x · x + 1 x. When multiplying fractions we have: 1 cos x · x + 1 x = x + 1 cos x · x. Having made all the substitutions, we get 1 - x cos x - x + 1 cos x · x. Now you need to work with fractions that have different denominators. We get:

x · 1 - x cos x · x - x + 1 cos x · x = x · 1 - x - 1 + x cos x · x = = x - x - x - 1 cos x · x = - x + 1 cos x x

Answer: 1 - x cos x - 1 c o s x · 1 + 1 x = - x + 1 cos x · x .

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Multiplying and dividing fractions.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

This operation is much nicer than addition-subtraction! Because it's easier. As a reminder, to multiply a fraction by a fraction, you need to multiply the numerators (this will be the numerator of the result) and the denominators (this will be the denominator). That is:

For example:

Everything is extremely simple. And please don't look for a common denominator! There is no need for him here...

To divide a fraction by a fraction, you need to reverse second(this is important!) fraction and multiply them, i.e.:

For example:

If you come across multiplication or division with integers and fractions, it’s okay. As with addition, we make a fraction from a whole number with one in the denominator - and go ahead! For example:

In high school, you often have to deal with three-story (or even four-story!) fractions. For example:

How can I make this fraction look decent? Yes, very simple! Use two-point division:

But don't forget about the order of division! Unlike multiplication, this is very important here! Of course, we will not confuse 4:2 or 2:4. But it’s easy to make a mistake in a three-story fraction. Please note for example:

In the first case (expression on the left):

In the second (expression on the right):

Do you feel the difference? 4 and 1/9!

What determines the order of division? Either with brackets, or (as here) with the length of horizontal lines. Develop your eye. And if there are no brackets or dashes, like:

then divide and multiply in order, from left to right!

And another very simple and important technique. In actions with degrees, it will be so useful to you! Let's divide one by any fraction, for example, by 13/15:

The shot has turned over! And this always happens. When dividing 1 by any fraction, the result is the same fraction, only upside down.

That's it for operations with fractions. The thing is quite simple, but it gives more than enough errors. Note practical advice, and there will be fewer of them (errors)!

Practical tips:

1. The most important thing when working with fractional expressions is accuracy and attentiveness! Is not common words, not good wishes! This is a dire necessity! Do all calculations on the Unified State Exam as a full-fledged task, focused and clear. It’s better to write two extra lines in your draft than to mess up when doing mental calculations.

2. In examples with different types fractions - go to ordinary fractions.

3. We reduce all fractions until they stop.

4. We reduce multi-level fractional expressions to ordinary ones using division through two points (we follow the order of division!).

5. Divide a unit by a fraction in your head, simply turning the fraction over.

Here are the tasks that you must definitely complete. Answers are given after all tasks. Use the materials on this topic and practical tips. Estimate how many examples you were able to solve correctly. The first time! Without a calculator! And draw the right conclusions...

Remember - the correct answer is received from the second (especially the third) time does not count! Such is the harsh life.

So, solve in exam mode ! This is already preparation for the Unified State Exam, by the way. We solve the example, check it, solve the next one. We decided everything - checked again from first to last. But only Then look at the answers.

Calculate:

Have you decided?

We are looking for answers that match yours. I deliberately wrote them down in disarray, away from temptation, so to speak... Here they are, the answers, written with semicolons.

0; 17/22; 3/4; 2/5; 1; 25.

Now we draw conclusions. If everything worked out, I’m happy for you! Basic calculations with fractions are not your problem! You can do more serious things. If not...

So you have one of two problems. Or both at once.) Lack of knowledge and (or) inattention. But this solvable Problems.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

To express a part as a fraction of the whole, you need to divide the part into the whole.

Task 1. There are 30 students in the class, four are absent. What proportion of students are absent?

Solution:

Answer: There are no students in the class.

Finding a fraction from a number

To solve problems in which you need to find a part of a whole that is fair next rule:

If a part of a whole is expressed as a fraction, then to find this part, you can divide the whole by the denominator of the fraction and multiply the result by its numerator.

Task 1. There were 600 rubles, this amount was spent. How much money did you spend?

Solution: to find 600 rubles or more, we need to divide this amount into 4 parts, thereby we will find out how much money one fourth part is:

600: 4 = 150 (r.)

Answer: spent 150 rubles.

Task 2. There were 1000 rubles, this amount was spent. How much money was spent?

Solution: from the problem statement we know that 1000 rubles consists of five equal parts. First, let’s find how many rubles are one-fifth of 1000, and then we’ll find out how many rubles are two-fifths:

1) 1000: 5 = 200 (r.) - one fifth.

2) 200 · 2 = 400 (r.) - two fifths.

These two actions can be combined: 1000: 5 · 2 = 400 (r.).

Answer: 400 rubles were spent.

The second way to find a part of a whole:

To find a part of a whole, you can multiply the whole by the fraction expressing that part of the whole.

Task 3. According to the charter of the cooperative, for the reporting meeting to be valid, at least at least members of the organization must be present. The cooperative has 120 members. What composition can a reporting meeting take place?

Solution:

Answer: the reporting meeting can take place if there are 80 members of the organization.

Finding a number by its fraction

To solve problems in which you need to find a whole from its part, the following rule applies:

If part of the desired whole is expressed as a fraction, then to find this whole, you can divide this part by the numerator of the fraction and multiply the result by its denominator.

Task 1. We spent 50 rubles, which was less than the original amount. Find the original amount of money.

Solution: from the description of the problem we see that 50 rubles is 6 times less than the original amount, i.e. the original amount is 6 times more than 50 rubles. To find this amount, you need to multiply 50 by 6:

50 · 6 = 300 (r.)

Answer: the initial amount is 300 rubles.

Task 2. We spent 600 rubles, which was less than the original amount of money. Find the original amount.

Solution: We will assume that the required number consists of three thirds. According to the condition, two-thirds of the number equals 600 rubles. First, let's find one third of the original amount, and then how many rubles are three thirds (the original amount):

1) 600: 2 3 = 900 (r.)

Answer: the initial amount is 900 rubles.

The second way to find a whole from its part:

To find a whole by the value expressing its part, you can divide this value by the fraction expressing this part.

Task 3. Line segment AB, equal to 42 cm, is the length of the segment CD. Find the length of the segment CD.

Solution:

Answer: segment length CD 70 cm.

Task 4. Watermelons were brought to the store. Before lunch, the store sold the watermelons it brought, and after lunch, there were 80 watermelons left to sell. How many watermelons did you bring to the store?

Solution: First, let’s find out what part of the brought watermelons is the number 80. To do this, let’s take the total number of watermelons brought as one and subtract from it the number of watermelons that were sold (sold):

And so, we learned that 80 watermelons make up the total number of watermelons brought. Now we find out how many watermelons from the total amount make up, and then how many watermelons make up (the number of watermelons brought):

2) 80: 4 15 = 300 (watermelons)

Answer: In total, 300 watermelons were brought to the store.

Let’s agree that “actions with fractions” in our lesson will mean actions with ordinary fractions. A common fraction is a fraction that has attributes such as a numerator, a fraction line, and a denominator. This distinguishes a common fraction from a decimal, which is obtained from a common fraction by reducing the denominator to a multiple of 10. Decimal written with a comma separating the whole part from the fractional part. We will talk about operations with ordinary fractions, since they are the ones that cause the greatest difficulties for students who have forgotten the basics of this topic, covered in the first half of the school mathematics course. At the same time, when transforming expressions into higher mathematics It is mainly actions with ordinary fractions that are used. The fraction abbreviations alone are worth it! Decimal fractions do not cause any particular difficulties. So, go ahead!

Two fractions are said to be equal if .

For example, since

Fractions and (since), and (since) are also equal.

Obviously, both fractions and are equal. This means that if the numerator and denominator of a given fraction are multiplied or divided by the same natural number, you will get a fraction equal to the given one: .

This property is called the basic property of a fraction.

The basic property of a fraction can be used to change the signs of the numerator and denominator of a fraction. If the numerator and denominator of a fraction are multiplied by -1, we get . This means that the value of a fraction will not change if the signs of the numerator and denominator are changed at the same time. If you change the sign of only the numerator or only the denominator, then the fraction will change its sign:

Reducing Fractions

Using the basic property of a fraction, you can replace a given fraction with another fraction that is equal to the given one, but with a smaller numerator and denominator. This substitution is called fraction reduction.

Let, for example, be given a fraction. The numbers 36 and 48 have a greatest common divisor of 12. Then

.

In general, reducing a fraction is always possible if the numerator and denominator are not mutually prime numbers. If the numerator and denominator are mutual prime numbers, then the fraction is called irreducible.

So, to reduce a fraction means to divide the numerator and denominator of the fraction by a common factor. All of the above also applies to fractional expressions containing variables.

Example 1. Reduce fraction

Solution. To factorize the numerator, first presenting the monomial - 5 xy as a sum - 2 xy - 3xy, we get

To factorize the denominator, we use the difference of squares formula:

As a result

.

Reducing fractions to a common denominator

Let two fractions and . They have different denominators: 5 and 7. Using the basic property of fractions, you can replace these fractions with others that are equal to them, and such that the resulting fractions will have the same denominators. Multiplying the numerator and denominator of the fraction by 7, we get

Multiplying the numerator and denominator of the fraction by 5, we get

So, the fractions are reduced to a common denominator:

.

But this is not the only solution to the problem: for example, these fractions can also be reduced to a common denominator of 70:

,

and in general to any denominator divisible by both 5 and 7.

Let's consider another example: let's bring the fractions and to a common denominator. Arguing as in the previous example, we get

,

.

But in this case, it is possible to reduce the fractions to a common denominator that is less than the product of the denominators of these fractions. Let's find the least common multiple of the numbers 24 and 30: LCM(24, 30) = 120.

Since 120:4 = 5, to write a fraction with a denominator of 120, you need to multiply both the numerator and the denominator by 5, this number is called an additional factor. Means .

Next, we get 120:30=4. Multiplying the numerator and denominator of the fraction by an additional factor of 4, we get .

So, these fractions are reduced to a common denominator.

The least common multiple of the denominators of these fractions is the smallest possible common denominator.

For fractional expressions that involve variables, the common denominator is a polynomial that is divided by the denominator of each fraction.

Example 2. Find the common denominator of the fractions and.

Solution. The common denominator of these fractions is a polynomial, since it is divisible by both and. However, this polynomial is not the only one that can be a common denominator of these fractions. It can also be a polynomial , and polynomial , and polynomial etc. Usually they take such a common denominator that any other common denominator is divided by the chosen one without a remainder. This denominator is called the lowest common denominator.

In our example, the lowest common denominator is . Got:

;

.

We were able to reduce fractions to their lowest common denominator. This happened by multiplying the numerator and denominator of the first fraction by , and the numerator and denominator of the second fraction by . Polynomials are called additional factors, respectively for the first and second fractions.

Adding and subtracting fractions

Addition of fractions is defined as follows:

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For example,

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If b = d, That

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This means that to add fractions with the same denominator, it is enough to add the numerators and leave the denominator the same. For example,

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If you add fractions with different denominators, you usually reduce the fractions to the lowest common denominator, and then add the numerators. For example,

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Now let's look at an example of adding fractional expressions with variables.

Example 3. Convert expression to one fraction

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Solution. Let's find the lowest common denominator. To do this, we first factorize the denominators.

Students are introduced to fractions in the 5th grade. Previously, people who knew how to perform operations with fractions were considered very smart. The first fraction was 1/2, that is, half, then 1/3 appeared, etc. For several centuries the examples were considered too complex. Now developed detailed rules on converting fractions, addition, multiplication and other operations. It is enough to understand the material a little, and the solution will be easy.

An ordinary fraction, called a simple fraction, is written as the division of two numbers: m and n.

M is the dividend, that is, the numerator of the fraction, and the divisor n is called the denominator.

Identify proper fractions (m< n) а также неправильные (m >n).

A proper fraction is less than one (for example, 5/6 - this means that 5 parts are taken from one; 2/8 - 2 parts are taken from one). An improper fraction is equal to or greater than 1 (8/7 - the unit is 7/7 and one more part is taken as a plus).

So, one is when the numerator and denominator coincide (3/3, 12/12, 100/100 and others).

Operations with ordinary fractions, grade 6

You can do the following with simple fractions:

• Expand a fraction. If you multiply the top and bottom part fractions for any same number(but not by zero), then the value of the fraction will not change (3/5 = 6/10 (simply multiplied by 2).
• Reducing fractions is similar to expanding, but here they divide by a number.
• Compare. If two fractions have the same numerators, then the fraction with the smaller denominator will be larger. If the denominators are the same, then the fraction with the largest numerator will be larger.
• Perform addition and subtraction. With the same denominators, this is easy to do (we sum up the upper parts, but the lower part does not change). If they are different, you will have to find a common denominator and additional factors.
• Multiply and divide fractions.

Let's look at examples of operations with fractions below.

Reduced fractions grade 6

To reduce is to divide the top and bottom of a fraction by some equal number.

The figure shows simple examples of reduction. In the first option, you can immediately guess that the numerator and denominator are divisible by 2.

On a note! If the number is even, then it is divisible by 2 in any way. Even numbers are 2, 4, 6...32 8 (ends with an even number), etc.

In the second case, when dividing 6 by 18, it is immediately clear that the numbers are divisible by 2. Dividing, we get 3/9. This fraction is further divided by 3. Then the answer is 1/3. If you multiply both divisors: 2 by 3, you get 6. It turns out that the fraction was divided by six. This gradual division is called successive reduction of fractions by common divisors.

Some people will immediately divide by 6, others will need to divide by parts. The main thing is that at the end there is a fraction left that cannot be reduced in any way.

Note that if a number consists of digits, the addition of which results in a number divisible by 3, then the original one can also be reduced by 3. Example: number 341. Add the numbers: 3 + 4 + 1 = 8 (8 is not divisible by 3, This means that the number 341 cannot be reduced by 3 without a remainder). Another example: 264. Add: 2 + 6 + 4 = 12 (divisible by 3). We get: 264: 3 = 88. This will make it easier to reduce large numbers.

In addition to the method of sequentially reducing fractions by common divisors, there are other methods.

GCD is the largest divisor for a number. Having found the gcd for the denominator and numerator, you can immediately reduce the fraction to the desired number. The search is carried out by gradually dividing each number. Next, they look at which divisors coincide; if there are several of them (as in the picture below), then you need to multiply.

Mixed Fractions Grade 6

All improper fractions can be converted into mixed fractions by separating the whole part from them. The whole number is written on the left.

Often comes from improper fraction do mixed number. The conversion process is shown in the example below: 22/4 = 22 divided by 4, we get 5 integers (5 * 4 = 20). 22 - 20 = 2. We get 5 integers and 2/4 (the denominator does not change). Since the fraction can be reduced, we divide the upper and lower parts by 2.

A mixed number can easily be converted into not correct fraction(this is necessary when dividing and multiplying fractions). To do this: multiply the integer by the lower part of the fraction and add the numerator to it. Ready. The denominator does not change.

Calculations with fractions 6th grade

Mixed numbers can be added. If the denominators are the same, then this is easy to do: add the integer parts and numerators, the denominator remains in place.

When adding numbers with different denominators, the process is more complicated. First, we reduce the numbers to one smallest denominator (LSD).

In the example below, for the numbers 9 and 6, the denominator will be 18. After this, additional factors are needed. To find them, you should divide 18 by 9, this is how you find the additional number - 2. We multiply it by the numerator 4 to get the fraction 8/18). They do the same with the second fraction. We already add the converted fractions (integers and numerators separately, we do not change the denominator). In the example, the answer had to be converted to a proper fraction (initially the numerator turned out to be greater than the denominator).

Please note that when fractions differ, the algorithm of actions is the same.

When multiplying fractions, it is important to place both under the same line. If the number is mixed, then we turn it into simple fraction. Next, multiply the upper and lower parts and write down the answer. If it is clear that fractions can be reduced, then we reduce them immediately.

In the above example, you didn’t have to cut anything, you just wrote down the answer and highlighted the whole part.

In this example, we had to reduce the numbers under one line. Although you can shorten the ready-made answer.

When dividing, the algorithm is almost the same. First, we turn the mixed fraction into an improper fraction, then we write the numbers under one line, replacing division with multiplication. Don’t forget to swap the top and bottom parts of the second fraction (this is the rule for dividing fractions).

If necessary, we reduce the numbers (in the example below we reduced them by five and two). We convert the improper fraction by highlighting the whole part.

Basic fraction problems 6th grade

The video shows a few more tasks. For clarity, graphic images of solutions are used to help visualize fractions.

Examples of multiplying fractions grade 6 with explanations

Multiplying fractions are written under one line. They are then reduced by dividing by the same numbers (for example, 15 in the denominator and 5 in the numerator can be divided by five).

Comparing fractions grade 6

To compare fractions, you need to remember two simple rules.

Rule 1. If the denominators are different

Rule 2. When the denominators are the same

For example, compare the fractions 7/12 and 2/3.

1. We look at the denominators, they do not match. So you need to find a common one.
2. For fractions, the common denominator is 12.
3. We first divide 12 by the lower part of the first fraction: 12: 12 = 1 (this is an additional factor for the 1st fraction).
4. Now we divide 12 by 3, we get 4 - extra. factor of the 2nd fraction.
5. We multiply the resulting numbers by the numerators to convert fractions: 1 x 7 = 7 (first fraction: 7/12); 4 x 2 = 8 (second fraction: 8/12).
6. Now we can compare: 7/12 and 8/12. It turned out: 7/12< 8/12.

To better represent fractions, you can use pictures for clarity where an object is divided into parts (for example, a cake). If you want to compare 4/7 and 2/3, then in the first case the cake is divided into 7 parts and 4 of them are selected. In the second, they divide into 3 parts and take 2. With the naked eye it will be clear that 2/3 will be greater than 4/7.

Examples with fractions grade 6 for training

You can complete the following tasks as practice.

• Compare fractions

• perform multiplication

Tip: if it is difficult to find the lowest common denominator for fractions (especially if their values ​​​​are small), then you can multiply the denominator of the first and second fractions. Example: 2/8 and 5/9. Finding their denominator is simple: multiply 8 by 9, you get 72.

Solving equations with fractions 6th grade

Solving equations requires remembering operations with fractions: multiplication, division, subtraction and addition. If one of the factors is unknown, then the product (total) is divided by the known factor, that is, the fractions are multiplied (the second is turned over).

If the dividend is unknown, then the denominator is multiplied by the divisor, and to find the divisor you need to divide the dividend by the quotient.

Let's imagine simple examples solutions to equations:

Here you only need to produce the difference of fractions, without leading to a common denominator.

The answer came out as an improper fraction. It can be converted to 1 whole and 3/5.

In the second method, the numerator and denominator were multiplied by 4 to cancel out the bottom portion rather than flipping the denominator.