To express a part as a fraction of the whole, you need to divide the part by the whole.
Task 1. There are 30 students in the class, four are missing. What proportion of students are missing?
Decision:
Answer: there are no students in the class.
Finding a fraction from a number
To solve problems in which it is required to find a part of the whole, it is true next rule:
If a part of the 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 have you spent?
Decision: to find from 600 rubles, you need to divide this amount into 4 parts, thereby we will find out how much money is one fourth:
600: 4 = 150 (p.)
Answer: spent 150 rubles.
Task 2. It was 1000 rubles, this amount was spent. How much money has been spent?
Decision: From the condition of the problem, we know that 1000 rubles consists of five equal parts. First we find how many rubles are one fifth of 1000, and then we find out how many rubles are two fifths:
1) 1000: 5 = 200 (p.) - one fifth.
2) 200 2 \u003d 400 (p.) - two fifths.
These two actions can be combined: 1000: 5 2 = 400 (p.).
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 a fraction expressing that part of the whole.
Task 3. According to the charter of the cooperative, for the validity of the reporting meeting, it must be attended by at least members of the organization. The cooperative has 120 members. With what composition can the reporting meeting be held?
Decision: 
Answer: the reporting meeting can be held if there are 80 members of the organization.
Finding a number by its fraction
To solve problems in which it is required to find the whole by its part, the following rule is true:
If a part of the desired integer is expressed as a fraction, then to find this integer, you can divide this part by the numerator of the fraction and multiply the result by its denominator.
Task 1. We spent 50 rubles, this amounted to the original amount. Find the original amount of money.
Decision: from the description of the problem, we see that 50 rubles is 6 times less than the initial amount, i.e., the initial 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, this amounted to the initial amount of money. Find the original amount.
Decision: we will assume that the desired number consists of three thirds. By condition, two-thirds of the number are equal to 600 rubles. First, we find one third of the initial amount, and then how many rubles are three-thirds (initial amount):
1) 600: 2 3 = 900 (p.)
Answer: the initial amount is 900 rubles.
The second way to find the whole by its part:
To find a whole by the value of its part, you can divide this value by a fraction that expresses this part.
Task 3. Line segment AB, equal to 42 cm, is the length of the segment CD. Find the length of a segment CD.
Decision: 
Answer: segment length CD 70 cm
Task 4. Watermelons were brought to the store. Before lunch, the store sold, after lunch - brought watermelons, and it remains to sell 80 watermelons. How many watermelons were brought to the store in total?
Decision: first, we find out what part of the imported watermelons is the number 80. To do this, we take the total number of imported watermelons as a unit and subtract from it the number of watermelons that we managed to sell (sell):
And so, we learned that 80 watermelons are from the total number of watermelons brought. Now we will find out how many watermelons of the total amount is, and then how many watermelons are (the number of watermelons brought):
2) 80: 4 15 = 300 (watermelons)
Answer: in total, 300 watermelons were brought to the store.
Convenient and simple online calculator fractions with detailed solution maybe:
- Add, subtract, multiply and divide fractions online,
- Receive turnkey solution fractions with a picture and it is convenient to transfer it.
The result of solving fractions will be here ...
0
1
2
3
4
5
6
7
8
9
Fraction sign "/" + - * :
_wipe Clear
Our online fraction calculator has fast input. To get the solution of fractions, for example, just write 1/2+2/7
into the calculator and press the " solve fractions". The calculator will write you detailed solution fractions and issue copy-friendly image.
The characters used for writing in the calculator
You can type an example for a solution both from the keyboard and using the buttons.
Features of the online fraction calculator
The fraction calculator can only perform operations with 2 simple fractions. They can be either correct (the numerator is less than the denominator) or incorrect (the numerator is greater than the denominator). The numbers in the numerator and denominators cannot be negative and greater than 999.Our online calculator solves fractions and brings the answer to correct form- reduces the fraction and highlights the whole part, if necessary.
If you need to solve negative fractions, just use the minus properties. When multiplying and dividing negative fractions, minus by minus gives plus. That is, the product and division of negative fractions is equal to the product and division of the same positive ones. If one fraction is negative when multiplied or divided, then simply remove the minus, and then add it to the answer. When adding negative fractions, the result will be the same as if you added the same positive fractions. If you add one negative fraction, then this is the same as subtracting the same positive one.
When subtracting negative fractions, the result will be the same as if they were reversed and made positive. That is minus by minus in this case gives a plus, and the sum does not change from the rearrangement of the terms. We use the same rules when subtracting fractions, one of which is negative.
To solve mixed fractions (fractions in which the whole part is highlighted), simply drive the whole part into a fraction. To do this, multiply the integer part by the denominator and add to the numerator.
If you need to solve 3 or more fractions online, then you should solve them one by one. First, count the first 2 fractions, then solve the next fraction with the answer received, and so on. Perform operations in turn for 2 fractions, and in the end you will get the correct answer.
Actions with fractions.
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
So, what are fractions, types of fractions, transformations - we remembered. Let's tackle the main question.
What can you do with fractions? Yes, everything is the same as with ordinary numbers. Add, subtract, multiply, divide.
All these actions with decimal operations with fractions are no different from operations with integers. Actually, this is what they are good for, decimal. The only thing is that you need to put the comma correctly.
mixed numbers , as I said, are of little use for most actions. They still need to be converted to ordinary fractions.
And here are the actions with ordinary fractions will be smarter. And much more important! Let me remind you: all actions with fractional expressions with letters, sines, unknowns, and so on and so forth are no different from actions with ordinary fractions! Operations with ordinary fractions are the basis for all algebra. It is for this reason that we will analyze all this arithmetic in great detail here.
Addition and subtraction of fractions.
Everyone can add (subtract) fractions with the same denominators (I really hope!). Well, let me remind you that I’m completely forgetful: when adding (subtracting), the denominator does not change. The numerators are added (subtracted) to give the numerator of the result. Type:
In short, in general terms:
What if the denominators are different? Then, using the main property of the fraction (here it came in handy again!), We make the denominators the same! For example:
Here we had to make the fraction 4/10 from the fraction 2/5. Solely for the purpose of making the denominators the same. I note, just in case, that 2/5 and 4/10 are the same fraction! Only 2/5 is uncomfortable for us, and 4/10 is even nothing.
By the way, this is the essence of solving any tasks in mathematics. When we're out uncomfortable expressions do the same, but more convenient to solve.
Another example:
The situation is similar. Here we make 48 out of 16. By simple multiplication on 3. This is all clear. But here we come across something like:
How to be?! It's hard to make a nine out of a seven! But we are smart, we know the rules! Let's transform every fraction so that the denominators are the same. This is called "reduce to a common denominator":
How! How did I know about 63? Very simple! 63 is a number that is evenly divisible by 7 and 9 at the same time. Such a number can always be obtained by multiplying the denominators. If we multiply some number by 7, for example, then the result will certainly be divided by 7!
If you need to add (subtract) several fractions, there is no need to do it in pairs, step by step. You just need to find the denominator that is common to all fractions, and bring each fraction to this same denominator. For example:
And what will be the common denominator? You can, of course, multiply 2, 4, 8, and 16. We get 1024. Nightmare. It is easier to estimate that the number 16 is perfectly divisible by 2, 4, and 8. Therefore, it is easy to get 16 from these numbers. This number will be the common denominator. Let's turn 1/2 into 8/16, 3/4 into 12/16, and so on.
By the way, if we take 1024 as a common denominator, everything will work out too, in the end everything will be reduced. Only not everyone will get to this end, because of the calculations ...
Solve the example yourself. Not a logarithm... It should be 29/16.
So, with the addition (subtraction) of fractions is clear, I hope? Of course, it is easier to work in a shortened version, with additional multipliers. But this pleasure is available to those who honestly worked in the lower grades ... And did not forget anything.
And now we will do the same actions, but not with fractions, but with fractional expressions. New rakes will be found here, yes ...
So, we need to add two fractional expressions:
We need to make the denominators the same. And only with the help multiplication! So the main property of the fraction says. Therefore, I cannot add one to x in the first fraction in the denominator. (But that would be nice!). But if you multiply the denominators, you see, everything will grow together! So we write down, the line of the fraction, leave an empty space on top, then add it, and write the product of the denominators below, so as not to forget:
And, of course, we don’t multiply anything on the right side, we don’t open brackets! And now, looking at the common denominator of the right side, we think: in order to get the denominator x (x + 1) in the first fraction, we need to multiply the numerator and denominator of this fraction by (x + 1). And in the second fraction - x. You get this:
Note! Parentheses are here! This is the rake that many step on. Not brackets, of course, but their absence. Parentheses appear because we multiply the whole numerator and the whole denominator! And not their individual pieces ...
In the numerator of the right side, we write the sum of the numerators, everything is as in numerical fractions, then we open the brackets in the numerator of the right side, i.e. multiply everything and give like. You don't need to open the brackets in the denominators, you don't need to multiply something! In general, in denominators (any) the product is always more pleasant! We get:
Here we got the answer. The process seems long and difficult, but it depends on practice. Solve examples, get used to it, everything will become simple. Those who have mastered the fractions in the allotted time, do all these operations with one hand, on the machine!
And one more note. Many famously deal with fractions, but hang on examples with whole numbers. Type: 2 + 1/2 + 3/4= ? Where to fasten a deuce? No need to fasten anywhere, you need to make a fraction out of a deuce. It's not easy, it's very simple! 2=2/1. Like this. Any whole number can be written as a fraction. The numerator is the number itself, the denominator is one. 7 is 7/1, 3 is 3/1 and so on. It's the same with letters. (a + b) \u003d (a + b) / 1, x \u003d x / 1, etc. And then we work with these fractions according to all the rules.
Well, on addition - subtraction of fractions, knowledge was refreshed. Transformations of fractions from one type to another - repeated. You can also check. Shall we settle a little?)
Calculate:
Answers (in disarray):
71/20; 3/5; 17/12; -5/4; 11/6
Multiplication / division of fractions - in the next lesson. There are also tasks for all actions with fractions.
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. Learning - with interest!)
you can get acquainted with functions and derivatives.
This article deals with 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, numeric expressions or expressions with variables. In conclusion, examples of solutions with a detailed description will be considered.
Yandex.RTB R-A-339285-1
Rules for performing operations with numerical fractions of a general form
Numerical fractions general view have a numerator and a denominator that contain integers or numeric expressions. If we consider such fractions 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 a different plan.
Definition 1
There are rules by which actions are performed with ordinary fractions. It is also suitable for fractions of a general form:
- When subtracting fractions with the same denominators, only the numerators are added, and the denominator remains the same, namely: a d ± c d \u003d a ± c d, the values a, c and d ≠ 0 are some numbers or numerical expressions.
- When adding or subtracting fractions with different denominators, it is necessary to reduce to a common one, and then add or subtract the resulting fractions with the same indicators. 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, an action is performed with numerators, after which with denominators, then we get a b c d \u003d 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 reciprocal, that is, we swap the numerator and denominator: a b: c d \u003d a b d c.
Rationale for the rules
Definition 2There are the following mathematical points that you should rely on when calculating:
- a fractional bar means a division sign;
- division by a number is treated as a multiplication by its reciprocal;
- application of the property of actions with real numbers;
- application of the basic property of a fraction and numerical inequalities.
With their help, you can make 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 actions with fractions. It is after this that the fraction needs to be simplified. This topic was discussed in detail in the section on converting fractions.
First, consider the example of adding and subtracting fractions with the same denominator.
Example 1
Given fractions 8 2 , 7 and 1 2 , 7 , then according to the rule it is necessary to add the numerator and rewrite the denominator.
Decision
Then we get a fraction of the form 8 + 1 2 , 7 . After performing the addition, we get 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 way to solve. To begin with, a transition is made 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 us subtract from 1 - 2 3 log 2 3 log 2 5 + 1 fractions 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 the reduction to a common denominator. Without this, we will not be able to further actions with fractions.
The process is remotely reminiscent of reduction to a common denominator. That is, a search is made for the least common divisor in the denominator, after which the missing factors are added to the fractions.
If the added fractions have no common factors, then their product can become one.
Example 3
Consider the example of adding fractions 2 3 5 + 1 and 1 2 .
Decision
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 to the first fraction it is equal to 2, and to the second 3 5 + 1. After multiplication, the fractions are reduced to the form 4 2 3 5 + 1. The general cast 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 fractions of a general form, then the least common denominator is usually not the case. It is unprofitable to take the product of numerators as a denominator. First you need to check if there is a number that is less in value than their product.
Example 4
Consider the example 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 a common denominator.
Consider examples of multiplications of fractions of a general form.
Example 5
To do this, it is necessary to multiply 2 + 1 6 and 2 · 5 3 · 2 + 1.
Decision
Following the rule, it is necessary to rewrite and write the product of numerators as a denominator. We get that 2 + 1 6 2 5 3 2 + 1 2 + 1 2 5 6 3 2 + 1 . When the fraction is multiplied, reductions can be made to simplify it. Then 5 3 3 2 + 1: 10 9 3 = 5 3 3 2 + 1 9 3 10 .
Using the rule of passage from division to multiplication by reciprocal, we get the reciprocal of the given one. To do this, the numerator and denominator are reversed. Let's look at an example:
5 3 3 2 + 1: 10 9 3 = 5 3 3 2 + 1 9 3 10
After that, they must perform multiplication and simplify the resulting fraction. If necessary, get rid of the 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 record will look like a multiplication of two fractions of the form 1 6 7 4 - 1 3 = 1 6 7 4 - 1 3 1 .
Performing an action with 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 valid 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 d0. A substitution of the form A D ± C D results in a difference of the form a 0 d 0 ± c 0 d 0 , where, according to 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 this we conclude that the chosen 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 to be a provable equality of the form A D ± C D = A ± C D .
Examples of addition and subtraction of fractions with variables
When there are the same denominators, it is only necessary 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 .
Decision
- To make a calculation, you need to subtract fractions that have the same denominators. Then we get that x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + 1 - 5 - x x + x - 2 . After that, you can open the brackets with the reduction of 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
- Since the denominators are the same, it remains only 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 the fraction can be reduced. Its numerator can be folded using the sum square formula, then we get (l g x + 2) 2 from the 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 - Given fractions of the form x - 1 x - 1 + x x + 1 with different denominators. After the transformation, you can proceed to addition.
Let's consider a two way solution.
The first method is that the denominator of the first fraction is subjected to factorization using squares, and 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 way is to multiply the numerator and denominator of the second fraction by x - 1 . Thus, we get rid of irrationality and proceed to adding a fraction 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. To add or subtract, you always need to look for a common denominator, which looks like the product of the denominators with the addition of additional factors to the numerators.
Example 7
Calculate the values of 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
Decision
- The denominator does not require any complicated calculations, so you need to choose their product of the form 3 x 7 + 2 2, then to the first fraction x 7 + 2 2 is chosen as an additional factor, and 3 to 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
- It can be seen that the denominators are presented as a product, which means that additional transformations are unnecessary. The common denominator will be the product of the form x 5 · ln 2 x + 1 · 2 x - 4 . From here 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) ) - This example makes sense when working with denominators of fractions. 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 pass 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 + x2
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 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.
Decision
You need to do the multiplication. 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 transferred 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 this 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 consider the action with fractions of a general form with exponentiation. If you have a degree natural indicator, then the action is considered as a multiplication of identical fractions. But it is recommended to use a general approach based on the properties of powers. 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 true. 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
The order of operations with fractions
Actions on fractions are performed according to certain rules. In practice, we notice that an expression can contain several fractions or fractional expressions. Then it is necessary to perform all actions in a strict order: raise to a power, multiply, divide, then add and subtract. If there are brackets, the first action is performed in them.
Example 9
Calculate 1 - x cos x - 1 c o s x · 1 + 1 x .
Decision
Since we have the same denominator, then 1 - x cos x and 1 c o s x , but it is impossible to subtract according to the rule, first the actions in brackets are performed, after which the multiplication, and then the 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 .
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
Lesson contentAdding fractions with the same denominators
Adding fractions is of two types:
- Adding fractions with the same denominators
- Adding fractions with different denominators
Let's start with adding fractions with the same denominators. Everything is simple here. To add fractions with the same denominators, you need to add their numerators, and leave the denominator unchanged. For example, let's add the fractions and . We add the numerators, and leave the denominator unchanged:
This example can be easily understood if we think of a pizza that is divided into four parts. If you add pizza to pizza, you get pizza:
Example 2 Add fractions and .
The answer turned out not proper fraction. If the end of the task comes, then it is customary to get rid of improper fractions. To get rid of an improper fraction, you need to select the whole part in it. In our case, the integer part is allocated easily - two divided by two is equal to one:
This example can be easily understood if we think of a pizza that is divided into two parts. If you add more pizzas to the pizza, you get one whole pizza:
Example 3. Add fractions and .
Again, add the numerators, and leave the denominator unchanged:
This example can be easily understood if we think of a pizza that is divided into three parts. If you add more pizzas to pizza, you get pizzas:
Example 4 Find the value of an expression 
This example is solved in exactly the same way as the previous ones. The numerators must be added and the denominator left unchanged:
Let's try to depict our solution using a picture. If you add pizzas to a pizza and add more pizzas, you get 1 whole pizza and more pizzas.
As you can see, adding fractions with the same denominators is not difficult. It is enough to understand the following rules:
- To add fractions with the same denominator, you need to add their numerators, and leave the denominator unchanged;
Adding fractions with different denominators
Now we will learn how to add fractions with different denominators. When adding fractions, the denominators of those fractions must be the same. But they are not always the same.
For example, fractions can be added because they have the same denominators.
But fractions cannot be added at once, because these fractions have different denominators. In such cases, fractions must be reduced to the same (common) denominator.
There are several ways to reduce fractions to the same denominator. Today we will consider only one of them, since the rest of the methods may seem complicated for a beginner.
The essence of this method lies in the fact that first (LCM) of the denominators of both fractions is sought. Then the LCM is divided by the denominator of the first fraction and the first additional factor is obtained. They do the same with the second fraction - the LCM is divided by the denominator of the second fraction and the second additional factor is obtained.
Then the numerators and denominators of the fractions are multiplied by their additional factors. As a result of these actions, fractions that had different denominators turn into fractions that have the same denominators. And we already know how to add such fractions.
Example 1. Add fractions and
First of all, we find the least common multiple of the denominators of both fractions. The denominator of the first fraction is the number 3, and the denominator of the second fraction is the number 2. The least common multiple of these numbers is 6
LCM (2 and 3) = 6
Now back to fractions and . First, we divide the LCM by the denominator of the first fraction and get the first additional factor. LCM is the number 6, and the denominator of the first fraction is the number 3. Divide 6 by 3, we get 2.
The resulting number 2 is the first additional factor. We write it down to the first fraction. To do this, we make a small oblique line above the fraction and write down the found additional factor above it:
We do the same with the second fraction. We divide the LCM by the denominator of the second fraction and get the second additional factor. LCM is the number 6, and the denominator of the second fraction is the number 2. Divide 6 by 2, we get 3.
The resulting number 3 is the second additional factor. We write it to the second fraction. Again, we make a small oblique line above the second fraction and write the found additional factor above it:
Now we are all set to add. It remains to multiply the numerators and denominators of fractions by their additional factors:
Look closely at what we have come to. We came to the conclusion that fractions that had different denominators turned into fractions that had the same denominators. And we already know how to add such fractions. Let's complete this example to the end:
Thus the example ends. To add it turns out.
Let's try to depict our solution using a picture. If you add pizzas to a pizza, you get one whole pizza and another sixth of a pizza:
Reduction of fractions to the same (common) denominator can also be depicted using a picture. Bringing the fractions and to a common denominator, we get the fractions and . These two fractions will be represented by the same slices of pizzas. The only difference will be that this time they will be divided into equal shares (reduced to the same denominator).
The first drawing shows a fraction (four pieces out of six) and the second picture shows a fraction (three pieces out of six). Putting these pieces together we get (seven pieces out of six). This fraction is incorrect, so we have highlighted the integer part in it. The result was (one whole pizza and another sixth pizza).
Note that we have painted given example too detailed. AT educational institutions it is not customary to write in such a detailed manner. You need to be able to quickly find the LCM of both denominators and additional factors to them, as well as quickly multiply the additional factors found by your numerators and denominators. While at school, we would have to write this example as follows:
But there is also the other side of the coin. If detailed notes are not made at the first stages of studying mathematics, then questions of the kind “Where does that number come from?”, “Why do fractions suddenly turn into completely different fractions? «.
To make it easier to add fractions with different denominators, you can use the following step-by-step instructions:
- Find the LCM of the denominators of fractions;
- Divide the LCM by the denominator of each fraction and get an additional multiplier for each fraction;
- Multiply the numerators and denominators of fractions by their additional factors;
- Add fractions that have the same denominators;
- If the answer turned out improper fraction, then select its integer part;
Example 2 Find the value of an expression 
.
Let's use the instructions above.
Step 1. Find the LCM of the denominators of fractions
Find the LCM of the denominators of both fractions. The denominators of the fractions are the numbers 2, 3 and 4
Step 2. Divide the LCM by the denominator of each fraction and get an additional multiplier for each fraction
Divide the LCM by the denominator of the first fraction. LCM is the number 12, and the denominator of the first fraction is the number 2. Divide 12 by 2, we get 6. We got the first additional factor 6. We write it over the first fraction:
Now we divide the LCM by the denominator of the second fraction. LCM is the number 12, and the denominator of the second fraction is the number 3. We divide 12 by 3, we get 4. We got the second additional factor 4. We write it over the second fraction:
Now we divide the LCM by the denominator of the third fraction. LCM is the number 12, and the denominator of the third fraction is the number 4. Divide 12 by 4, we get 3. We got the third additional factor 3. We write it over the third fraction:
Step 3. Multiply the numerators and denominators of fractions by your additional factors
We multiply the numerators and denominators by our additional factors:
Step 4. Add fractions that have the same denominators
We came to the conclusion that fractions that had different denominators turned into fractions that have the same (common) denominators. It remains to add these fractions. Add up:
The addition didn't fit on one line, so we moved the remaining expression to the next line. This is allowed in mathematics. When an expression does not fit on one line, it is carried over to the next line, and it is necessary to put an equal sign (=) at the end of the first line and at the beginning new line. The equal sign on the second line indicates that this is a continuation of the expression that was on the first line.
Step 5. If the answer turned out to be an improper fraction, then select the whole part in it
Our answer is an improper fraction. We must single out the whole part of it. We highlight:
Got an answer
Subtraction of fractions with the same denominators
There are two types of fraction subtraction:
- Subtraction of fractions with the same denominators
- Subtraction of fractions with different denominators
First, let's learn how to subtract fractions with the same denominators. Everything is simple here. To subtract another from one fraction, you need to subtract the numerator of the second fraction from the numerator of the first fraction, and leave the denominator the same.
For example, let's find the value of the expression . To solve this example, it is necessary to subtract the numerator of the second fraction from the numerator of the first fraction, and leave the denominator unchanged. Let's do this:
This example can be easily understood if we think of a pizza that is divided into four parts. If you cut pizzas from a pizza, you get pizzas:
Example 2 Find the value of the expression .
Again, from the numerator of the first fraction, subtract the numerator of the second fraction, and leave the denominator unchanged:
This example can be easily understood if we think of a pizza that is divided into three parts. If you cut pizzas from a pizza, you get pizzas:
Example 3 Find the value of an expression 
This example is solved in exactly the same way as the previous ones. From the numerator of the first fraction, you need to subtract the numerators of the remaining fractions:
As you can see, there is nothing complicated in subtracting fractions with the same denominators. It is enough to understand the following rules:
- To subtract another from one fraction, you need to subtract the numerator of the second fraction from the numerator of the first fraction, and leave the denominator unchanged;
- If the answer turned out to be an improper fraction, then you need to select the whole part in it.
Subtraction of fractions with different denominators
For example, a fraction can be subtracted from a fraction, since these fractions have the same denominators. But a fraction cannot be subtracted from a fraction, because these fractions have different denominators. In such cases, fractions must be reduced to the same (common) denominator.
The common denominator is found according to the same principle that we used when adding fractions with different denominators. First of all, find the LCM of the denominators of both fractions. Then the LCM is divided by the denominator of the first fraction and the first additional factor is obtained, which is written over the first fraction. Similarly, the LCM is divided by the denominator of the second fraction and a second additional factor is obtained, which is written over the second fraction.
The fractions are then multiplied by their additional factors. As a result of these operations, fractions that had different denominators turn into fractions that have the same denominators. And we already know how to subtract such fractions.
Example 1 Find the value of an expression:
These fractions have different denominators, so you need to bring them to the same (common) denominator.
First, we find the LCM of the denominators of both fractions. The denominator of the first fraction is the number 3, and the denominator of the second fraction is the number 4. The least common multiple of these numbers is 12
LCM (3 and 4) = 12
Now back to fractions and
Let's find an additional factor for the first fraction. To do this, we divide the LCM by the denominator of the first fraction. LCM is the number 12, and the denominator of the first fraction is the number 3. Divide 12 by 3, we get 4. We write the four over the first fraction:
We do the same with the second fraction. We divide the LCM by the denominator of the second fraction. LCM is the number 12, and the denominator of the second fraction is the number 4. Divide 12 by 4, we get 3. Write a triple over the second fraction:
Now we are all set for subtraction. It remains to multiply the fractions by their additional factors:
We came to the conclusion that fractions that had different denominators turned into fractions that had the same denominators. And we already know how to subtract such fractions. Let's complete this example to the end:
Got an answer
Let's try to depict our solution using a picture. If you cut pizzas from a pizza, you get pizzas.
This is the detailed version of the solution. Being at school, we would have to solve this example in a shorter way. Such a solution would look like this:
Reduction of fractions and to a common denominator can also be depicted using a picture. Bringing these fractions to a common denominator, we get the fractions and . These fractions will be represented by the same pizza slices, but this time they will be divided into the same fractions (reduced to the same denominator):
The first drawing shows a fraction (eight pieces out of twelve), and the second picture shows a fraction (three pieces out of twelve). By cutting off three pieces from eight pieces, we get five pieces out of twelve. The fraction describes these five pieces.
Example 2 Find the value of an expression 
These fractions have different denominators, so you first need to bring them to the same (common) denominator.
Find the LCM of the denominators of these fractions.
The denominators of the fractions are the numbers 10, 3 and 5. The least common multiple of these numbers is 30
LCM(10, 3, 5) = 30
Now we find additional factors for each fraction. To do this, we divide the LCM by the denominator of each fraction.
Let's find an additional factor for the first fraction. LCM is the number 30, and the denominator of the first fraction is the number 10. Divide 30 by 10, we get the first additional factor 3. We write it over the first fraction:
Now we find an additional factor for the second fraction. Divide the LCM by the denominator of the second fraction. LCM is the number 30, and the denominator of the second fraction is the number 3. Divide 30 by 3, we get the second additional factor 10. We write it over the second fraction:
Now we find an additional factor for the third fraction. Divide the LCM by the denominator of the third fraction. LCM is the number 30, and the denominator of the third fraction is the number 5. Divide 30 by 5, we get the third additional factor 6. We write it over the third fraction:
Now everything is ready for subtraction. It remains to multiply the fractions by their additional factors:
We came to the conclusion that fractions that had different denominators turned into fractions that have the same (common) denominators. And we already know how to subtract such fractions. Let's finish this example.
The continuation of the example will not fit on one line, so we move the continuation to the next line. Don't forget about the equal sign (=) on the new line:
The answer turned out to be a correct fraction, and everything seems to suit us, but it is too cumbersome and ugly. We should make it easier. What can be done? You can reduce this fraction.
To reduce a fraction, you need to divide its numerator and denominator by (gcd) the numbers 20 and 30.
So, we find the GCD of the numbers 20 and 30:
Now we return to our example and divide the numerator and denominator of the fraction by the found GCD, that is, by 10
Got an answer
Multiplying a fraction by a number
To multiply a fraction by a number, you need to multiply the numerator of the given fraction by this number, and leave the denominator the same.
Example 1. Multiply the fraction by the number 1.
Multiply the numerator of the fraction by the number 1
The entry can be understood as taking half 1 time. For example, if you take pizza 1 time, you get pizza
From the laws of multiplication, we know that if the multiplicand and the multiplier are interchanged, then the product will not change. If the expression is written as , then the product will still be equal to . Again, the rule for multiplying an integer and a fraction works:
This entry can be understood as taking half of the unit. For example, if there is 1 whole pizza and we take half of it, then we will have pizza:
Example 2. Find the value of an expression
Multiply the numerator of the fraction by 4
The answer is an improper fraction. Let's take a whole part of it:
The expression can be understood as taking two quarters 4 times. For example, if you take pizzas 4 times, you get two whole pizzas.
And if we swap the multiplicand and the multiplier in places, we get the expression. It will also be equal to 2. This expression can be understood as taking two pizzas from four whole pizzas:
Multiplication of fractions
To multiply fractions, you need to multiply their numerators and denominators. If the answer is an improper fraction, you need to select the whole part in it.
Example 1 Find the value of the expression .
Got an answer. It is desirable to reduce given fraction. The fraction can be reduced by 2. Then the final solution will take the following form:
The expression can be understood as taking a pizza from half a pizza. Let's say we have half a pizza:
How to take two-thirds from this half? First you need to divide this half into three equal parts:
And take two from these three pieces:
We'll get pizza. Remember what a pizza looks like divided into three parts:
One slice from this pizza and the two slices we took will have the same dimensions:
In other words, we are talking about the same pizza size. Therefore, the value of the expression is
Example 2. Find the value of an expression
Multiply the numerator of the first fraction by the numerator of the second fraction, and the denominator of the first fraction by the denominator of the second fraction:
The answer is an improper fraction. Let's take a whole part of it:
Example 3 Find the value of an expression
Multiply the numerator of the first fraction by the numerator of the second fraction, and the denominator of the first fraction by the denominator of the second fraction:
The answer turned out to be a correct fraction, but it will be good if it is reduced. To reduce this fraction, you need to divide the numerator and denominator of this fraction by the largest common divisor(gcd) numbers 105 and 450.
So, let's find the GCD of the numbers 105 and 450:
Now we divide the numerator and denominator of our answer to the GCD that we have now found, that is, by 15
Representing an integer as a fraction
Any whole number can be represented as a fraction. For example, the number 5 can be represented as . From this, five will not change its meaning, since the expression means “the number five divided by one”, and this, as you know, is equal to five:
Reverse numbers
Now we will get acquainted with interesting topic in mathematics. It's called "reverse numbers".
Definition. Reverse to numbera is the number that, when multiplied bya gives a unit.
Let's substitute in this definition instead of a variable a number 5 and try to read the definition:
Reverse to number 5 is the number that, when multiplied by 5 gives a unit.
Is it possible to find a number that, when multiplied by 5, gives one? It turns out you can. Let's represent five as a fraction:
Then multiply this fraction by itself, just swap the numerator and denominator. In other words, let's multiply the fraction by itself, only inverted:
What will be the result of this? If we continue to solve this example, we get one:
This means that the inverse of the number 5 is the number, since when 5 is multiplied by one, one is obtained.
The reciprocal can also be found for any other integer.
You can also find the reciprocal for any other fraction. To do this, it is enough to turn it over.
Division of a fraction by a number
Let's say we have half a pizza:
Let's divide it equally between two. How many pizzas will each get?
It can be seen that after splitting half of the pizza, two equal pieces were obtained, each of which makes up a pizza. So everyone gets a pizza.
Division of fractions is done using reciprocals. Reverse numbers allow you to replace division with multiplication.
To divide a fraction by a number, you need to multiply this fraction by the reciprocal of the divisor.
Using this rule, we will write down the division of our half of the pizza into two parts.
So, you need to divide the fraction by the number 2. Here the dividend is a fraction and the divisor is 2.
To divide a fraction by the number 2, you need to multiply this fraction by the reciprocal of the divisor 2. The reciprocal of the divisor 2 is a fraction. So you need to multiply by
