The arithmetic operation that is performed last when calculating the value of an expression is the “master” operation.
That is, if you substitute some (any) numbers instead of letters and try to calculate the value of the expression, then if the last action is multiplication, then we have a product (the expression is factorized).
If the last action is addition or subtraction, this means that the expression is not factorized (and therefore cannot be reduced).
To reinforce this, solve a few examples yourself:
Examples:
Solutions:
1. I hope you didn’t immediately rush to cut and? It was still not enough to “reduce” units like this:
The first step should be factorization:
4. Adding and subtracting fractions. Reducing fractions to a common denominator.
Adding and subtracting ordinary fractions is a familiar operation: we look for a common denominator, multiply each fraction by the missing factor and add/subtract the numerators.
Let's remember:
Answers:
1. The denominators and are relatively prime, that is, they do not have common factors. Therefore, the LCM of these numbers is equal to their product. This will be the common denominator:
2. Here the common denominator is:
3. Here, first of all, we convert mixed fractions into improper ones, and then according to the usual scheme:
It's a completely different matter if the fractions contain letters, for example:
Let's start with something simple:
a) Denominators do not contain letters
Here everything is the same as with ordinary numerical fractions: we find the common denominator, multiply each fraction by the missing factor and add/subtract the numerators:
Now in the numerator you can give similar ones, if any, and factor them:
Try it yourself:
Answers:
b) Denominators contain letters
Let's remember the principle of finding a common denominator without letters:
· first of all, we determine the common factors;
· then we write out all the common factors one at a time;
· and multiply them by all other non-common factors.
To determine the common factors of the denominators, we first factor them into prime factors:
Let us emphasize the common factors:
Now let’s write out the common factors one at a time and add to them all the non-common (not underlined) factors:
This is the common denominator.
Let's get back to the letters. The denominators are given in exactly the same way:
· factor the denominators;
· determine common (identical) factors;
· write out all common factors once;
· multiply them by all other non-common factors.
So, in order:
1) factor the denominators:
2) determine common (identical) factors:
3) write out all common factors once and multiply them by all other (unemphasized) factors:
So there's a common denominator here. The first fraction must be multiplied by, the second - by:
By the way, there is one trick:
For example: .
We see the same factors in the denominators, only all with different indicators. The common denominator will be:
to a degree
to a degree
to a degree
to a degree.
Let's complicate the task:
How to make fractions have the same denominator?
Let's remember the basic property of a fraction:
Nowhere does it say that the same number can be subtracted (or added) from the numerator and denominator of a fraction. Because it's not true!
See for yourself: take any fraction, for example, and add some number to the numerator and denominator, for example, . What did you learn?
So, another unshakable rule:
When you reduce fractions to a common denominator, use only the multiplication operation!
But what do you need to multiply by to get?
So multiply by. And multiply by:
We will call expressions that cannot be factorized “elementary factors.”
For example, - this is an elementary multiplier. - Same. But no: it can be factorized.
What about the expression? Is it elementary?
No, because it can be factorized:
(you already read about factorization in the topic “”).
So, the elementary factors into which you decompose an expression with letters are an analogue of the simple factors into which you decompose numbers. And we will deal with them in the same way.
We see that both denominators have a multiplier. It will go to the common denominator to the degree (remember why?).
The factor is elementary, and they do not have a common factor, which means that the first fraction will simply have to be multiplied by it:
Another example:
Solution:
Before you multiply these denominators in a panic, you need to think about how to factor them? They both represent:
Great! Then:
Another example:
Solution:
As usual, let's factorize the denominators. In the first denominator we simply put it out of brackets; in the second - the difference of squares:
It would seem that there are no common factors. But if you look closely, they are similar... And it’s true:
So let's write:
That is, it turned out like this: inside the bracket we swapped the terms, and at the same time the sign in front of the fraction changed to the opposite. Take note, you will have to do this often.
Now let's bring it to a common denominator:
Got it? Let's check it now.
Tasks for independent solution:
Answers:
Here we need to remember one more thing - the difference of cubes:
Please note that the denominator of the second fraction does not contain the formula “square of the sum”! The square of the sum would look like this: .
A is the so-called incomplete square of the sum: the second term in it is the product of the first and last, and not their double product. The partial square of the sum is one of the factors in the expansion of the difference of cubes:
What to do if there are already three fractions?
Yes, the same thing! First of all, let’s make sure that the maximum number of factors in the denominators is the same:
Please note: if you change the signs inside one bracket, the sign in front of the fraction changes to the opposite. When we change the signs in the second bracket, the sign in front of the fraction changes again to the opposite. As a result, it (the sign in front of the fraction) has not changed.
We write out the entire first denominator into the common denominator, and then add to it all the factors that have not yet been written, from the second, and then from the third (and so on, if there are more fractions). That is, it turns out like this:
Hmm... It’s clear what to do with fractions. But what about the two?
It's simple: you know how to add fractions, right? So, we need to make two become a fraction! Let's remember: a fraction is a division operation (the numerator is divided by the denominator, in case you forgot). And there is nothing easier than dividing a number by. In this case, the number itself will not change, but will turn into a fraction:
Exactly what is needed!
5. Multiplication and division of fractions.
Well, the hardest part is over now. And ahead of us is the simplest, but at the same time the most important:
Procedure
What is the procedure for calculating a numerical expression? Remember by calculating the meaning of this expression:
Did you count?
It should work.
So, let me remind you.
The first step is to calculate the degree.
The second is multiplication and division. If there are several multiplications and divisions at the same time, they can be done in any order.
And finally, we perform addition and subtraction. Again, in any order.
But: the expression in brackets is evaluated out of turn!
If several brackets are multiplied or divided by each other, we first calculate the expression in each of the brackets, and then multiply or divide them.
What if there are more brackets inside the brackets? Well, let's think: some expression is written inside the brackets. When calculating an expression, what should you do first? That's right, calculate the brackets. Well, we figured it out: first we calculate the inner brackets, then everything else.
So, the procedure for the expression above is as follows (the current action is highlighted in red, that is, the action that I am performing right now):
Okay, it's all simple.
But this is not the same as an expression with letters?
No, it's the same! Only instead of arithmetic operations, you need to do algebraic ones, that is, the actions described in the previous section: bringing similar, adding fractions, reducing fractions and so on. The only difference will be the action of factoring polynomials (we often use this when working with fractions). Most often, to factorize, you need to use I or simply put the common factor out of brackets.
Usually our goal is to represent an expression as a product or quotient.
For example:
Let's simplify the expression.
1) First, we simplify the expression in brackets. There we have a difference of fractions, and our goal is to present it as a product or quotient. So, we bring the fractions to a common denominator and add:
It is impossible to simplify this expression any further; all the factors here are elementary (do you still remember what this means?).
2) We get:
Multiplying fractions: what could be simpler.
3) Now you can shorten:
OK it's all over Now. Nothing complicated, right?
Another example:
Simplify the expression.
First, try to solve it yourself, and only then look at the solution.
Solution:
First of all, let's determine the order of actions.
First, let's add the fractions in parentheses, so instead of two fractions we get one.
Then we will do division of fractions. Well, let's add the result with the last fraction.
I will number the steps schematically:
Now I’ll show you the process, tinting the current action in red:
1. If there are similar ones, they must be brought immediately. At whatever point similar ones arise in our country, it is advisable to bring them up immediately.
2. The same applies to reducing fractions: as soon as the opportunity to reduce appears, it must be taken advantage of. The exception is for fractions that you add or subtract: if they now have the same denominators, then the reduction should be left for later.
Here are some tasks for you to solve on your own:
And what was promised at the very beginning:
Answers:
Solutions (brief):
If you have coped with at least the first three examples, then you have mastered the topic.
Now on to learning!
CONVERTING EXPRESSIONS. SUMMARY AND BASIC FORMULAS
Basic simplification operations:
- Bringing similar: to add (reduce) similar terms, you need to add their coefficients and assign the letter part.
- Factorization: putting the common factor out of brackets, applying it, etc.
- Reducing a fraction: The numerator and denominator of a fraction can be multiplied or divided by the same non-zero number, which does not change the value of the fraction.
1) numerator and denominator factorize
2) if the numerator and denominator have common factors, they can be crossed out.IMPORTANT: only multipliers can be reduced!
- Adding and subtracting fractions:
; - Multiplying and dividing fractions:
;
Let's consider the topic of transforming expressions with powers, but first let's dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open parentheses, add similar terms, work with bases and exponents, and use the properties of powers.
What are power expressions?
In school courses, few people use the phrase “powerful expressions,” but this term is constantly found in collections for preparing for the Unified State Exam. In most cases, a phrase denotes expressions that contain degrees in their entries. This is what we will reflect in our definition.
Definition 1
Power expression is an expression that contains powers.
Let us give several examples of power expressions, starting with a power with a natural exponent and ending with a power with a real exponent.
The simplest power expressions can be considered powers of a number with a natural exponent: 3 2, 7 5 + 1, (2 + 1) 5, (− 0, 1) 4, 2 2 3 3, 3 a 2 − a + a 2, x 3 − 1 , (a 2) 3 . And also powers with zero exponent: 5 0, (a + 1) 0, 3 + 5 2 − 3, 2 0. And powers with negative integer powers: (0, 5) 2 + (0, 5) - 2 2.
It is a little more difficult to work with a degree that has rational and irrational exponents: 264 1 4 - 3 3 3 1 2, 2 3, 5 2 - 2 2 - 1, 5, 1 a 1 4 a 1 2 - 2 a - 1 6 · b 1 2 , x π · x 1 - π , 2 3 3 + 5 .
The indicator can be the variable 3 x - 54 - 7 3 x - 58 or the logarithm x 2 · l g x − 5 · x l g x.
We have dealt with the question of what power expressions are. Now let's start converting them.
Main types of transformations of power expressions
First of all, we will look at the basic identity transformations of expressions that can be performed with power expressions.
Example 1
Calculate the value of a power expression 2 3 (4 2 − 12).
Solution
We will carry out all transformations in compliance with the order of actions. In this case, we will start by performing the actions in brackets: we will replace the degree with a digital value and calculate the difference of two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.
All we have to do is replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here's our answer.
Answer: 2 3 · (4 2 − 12) = 32 .
Example 2
Simplify the expression with powers 3 a 4 b − 7 − 1 + 2 a 4 b − 7.
Solution
The expression given to us in the problem statement contains similar terms that we can give: 3 a 4 b − 7 − 1 + 2 a 4 b − 7 = 5 a 4 b − 7 − 1.
Answer: 3 · a 4 · b − 7 − 1 + 2 · a 4 · b − 7 = 5 · a 4 · b − 7 − 1 .
Example 3
Express the expression with powers 9 - b 3 · π - 1 2 as a product.
Solution
Let's imagine the number 9 as a power 3 2 and apply the abbreviated multiplication formula:
9 - b 3 π - 1 2 = 3 2 - b 3 π - 1 2 = = 3 - b 3 π - 1 3 + b 3 π - 1
Answer: 9 - b 3 · π - 1 2 = 3 - b 3 · π - 1 3 + b 3 · π - 1 .
Now let's move on to the analysis of identity transformations that can be applied specifically to power expressions.
Working with base and exponent
The degree in the base or exponent can have numbers, variables, and some expressions. For example, (2 + 0, 3 7) 5 − 3, 7 And . Working with such records is difficult. It is much easier to replace the expression in the base of the degree or the expression in the exponent with an identically equal expression.
Transformations of degree and exponent are carried out according to the rules known to us separately from each other. The most important thing is that the transformation results in an expression identical to the original one.
The purpose of transformations is to simplify the original expression or obtain a solution to the problem. For example, in the example we gave above, (2 + 0, 3 7) 5 − 3, 7 you can follow the steps to go to the degree 4 , 1 1 , 3 . By opening the parentheses, we can present similar terms to the base of the power (a · (a + 1) − a 2) 2 · (x + 1) and obtain a power expression of a simpler form a 2 (x + 1).
Using Degree Properties
Properties of powers, written in the form of equalities, are one of the main tools for transforming expressions with powers. We present here the main ones, taking into account that a And b are any positive numbers, and r And s- arbitrary real numbers:
Definition 2
- a r · a s = a r + s ;
- a r: a s = a r − s ;
- (a · b) r = a r · b r ;
- (a: b) r = a r: b r ;
- (a r) s = a r · s .
In cases where we are dealing with natural, integer, positive exponents, the restrictions on the numbers a and b can be much less strict. So, for example, if we consider the equality a m · a n = a m + n, Where m And n are natural numbers, then it will be true for any values of a, both positive and negative, as well as for a = 0.
The properties of powers can be used without restrictions in cases where the bases of the powers are positive or contain variables whose range of permissible values is such that the bases take only positive values on it. In fact, in the school mathematics curriculum, the student's task is to select an appropriate property and apply it correctly.
When preparing to enter universities, you may encounter problems in which inaccurate application of properties will lead to a narrowing of the DL and other difficulties in solving. In this section we will examine only two such cases. More information on the subject can be found in the topic “Converting expressions using properties of powers”.
Example 4
Imagine the expression a 2 , 5 (a 2) − 3: a − 5 , 5 in the form of a power with a base a.
Solution
First, we use the property of exponentiation and transform the second factor using it (a 2) − 3. Then we use the properties of multiplication and division of powers with the same base:
a 2 , 5 · a − 6: a − 5 , 5 = a 2 , 5 − 6: a − 5 , 5 = a − 3 , 5: a − 5 , 5 = a − 3 , 5 − (− 5 , 5) = a 2 .
Answer: a 2, 5 · (a 2) − 3: a − 5, 5 = a 2.
Transformation of power expressions according to the property of powers can be done both from left to right and in the opposite direction.
Example 5
Find the value of the power expression 3 1 3 · 7 1 3 · 21 2 3 .
Solution
If we apply equality (a · b) r = a r · b r, from right to left, we get a product of the form 3 · 7 1 3 · 21 2 3 and then 21 1 3 · 21 2 3 . Let's add the exponents when multiplying powers with the same bases: 21 1 3 · 21 2 3 = 21 1 3 + 2 3 = 21 1 = 21.
There is another way to carry out the transformation:
3 1 3 · 7 1 3 · 21 2 3 = 3 1 3 · 7 1 3 · (3 · 7) 2 3 = 3 1 3 · 7 1 3 · 3 2 3 · 7 2 3 = = 3 1 3 · 3 2 3 7 1 3 7 2 3 = 3 1 3 + 2 3 7 1 3 + 2 3 = 3 1 7 1 = 21
Answer: 3 1 3 7 1 3 21 2 3 = 3 1 7 1 = 21
Example 6
Given a power expression a 1, 5 − a 0, 5 − 6, enter a new variable t = a 0.5.
Solution
Let's imagine the degree a 1, 5 How a 0.5 3. Using the property of degrees to degrees (a r) s = a r · s from right to left and we get (a 0, 5) 3: a 1, 5 − a 0, 5 − 6 = (a 0, 5) 3 − a 0, 5 − 6. You can easily introduce a new variable into the resulting expression t = a 0.5: we get t 3 − t − 6.
Answer: t 3 − t − 6 .
Converting fractions containing powers
We usually deal with two versions of power expressions with fractions: the expression represents a fraction with a power or contains such a fraction. All basic transformations of fractions are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, or worked separately with the numerator and denominator. Let's illustrate this with examples.
Example 7
Simplify the power expression 3 · 5 2 3 · 5 1 3 - 5 - 2 3 1 + 2 · x 2 - 3 - 3 · x 2 .
Solution
We are dealing with a fraction, so we will carry out transformations in both the numerator and the denominator:
3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = 3 5 2 3 5 1 3 - 3 5 2 3 5 - 2 3 - 2 - x 2 = = 3 5 2 3 + 1 3 - 3 5 2 3 + - 2 3 - 2 - x 2 = 3 5 1 - 3 5 0 - 2 - x 2
Place a minus sign in front of the fraction to change the sign of the denominator: 12 - 2 - x 2 = - 12 2 + x 2
Answer: 3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = - 12 2 + x 2
Fractions containing powers are reduced to a new denominator in the same way as rational fractions. To do this, you need to find an additional factor and multiply the numerator and denominator of the fraction by it. It is necessary to select an additional factor in such a way that it does not go to zero for any values of variables from the ODZ variables for the original expression.
Example 8
Reduce the fractions to a new denominator: a) a + 1 a 0, 7 to the denominator a, b) 1 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 to the denominator x + 8 · y 1 2 .
Solution
a) Let's select a factor that will allow us to reduce to a new denominator. a 0, 7 a 0, 3 = a 0, 7 + 0, 3 = a, therefore, as an additional factor we will take a 0 , 3. The range of permissible values of the variable a includes the set of all positive real numbers. Degree in this field a 0 , 3 does not go to zero.
Let's multiply the numerator and denominator of a fraction by a 0 , 3:
a + 1 a 0, 7 = a + 1 a 0, 3 a 0, 7 a 0, 3 = a + 1 a 0, 3 a
b) Let's pay attention to the denominator:
x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 2 - x 1 3 2 y 1 6 + 2 y 1 6 2
Let's multiply this expression by x 1 3 + 2 · y 1 6, we get the sum of the cubes x 1 3 and 2 · y 1 6, i.e. x + 8 · y 1 2 . This is our new denominator to which we need to reduce the original fraction.
This is how we found the additional factor x 1 3 + 2 · y 1 6 . On the range of permissible values of variables x And y the expression x 1 3 + 2 y 1 6 does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:
1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 + 2 y 1 6 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 3 + 2 y 1 6 3 = x 1 3 + 2 y 1 6 x + 8 y 1 2
Answer: a) a + 1 a 0, 7 = a + 1 a 0, 3 a, b) 1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = x 1 3 + 2 y 1 6 x + 8 · y 1 2 .
Example 9
Reduce the fraction: a) 30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3, b) a 1 4 - b 1 4 a 1 2 - b 1 2.
Solution
a) We use the greatest common denominator (GCD), by which we can reduce the numerator and denominator. For numbers 30 and 45 it is 15. We can also make a reduction by x0.5+1 and on x + 2 · x 1 1 3 - 5 3 .
We get:
30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 x 3 3 (x 0, 5 + 1)
b) Here the presence of identical factors is not obvious. You will have to perform some transformations in order to get the same factors in the numerator and denominator. To do this, we expand the denominator using the difference of squares formula:
a 1 4 - b 1 4 a 1 2 - b 1 2 = a 1 4 - b 1 4 a 1 4 2 - b 1 2 2 = = a 1 4 - b 1 4 a 1 4 + b 1 4 a 1 4 - b 1 4 = 1 a 1 4 + b 1 4
Answer: a) 30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 · x 3 3 · (x 0 , 5 + 1) , b) a 1 4 - b 1 4 a 1 2 - b 1 2 = 1 a 1 4 + b 1 4 .
Basic operations with fractions include converting fractions to a new denominator and reducing fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, first the fractions are reduced to a common denominator, after which operations (addition or subtraction) are carried out with the numerators. The denominator remains the same. The result of our actions is a new fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators.
Example 10
Do the steps x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2 .
Solution
Let's start by subtracting the fractions that are in parentheses. Let's bring them to a common denominator:
x 1 2 - 1 x 1 2 + 1
Let's subtract the numerators:
x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 + 1 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 - x 1 2 - 1 x 1 2 - 1 x 1 2 + 1 x 1 2 - 1 1 x 1 2 = = x 1 2 + 1 2 - x 1 2 - 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 2 + 2 x 1 2 + 1 - x 1 2 2 - 2 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2
Now we multiply the fractions:
4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 x 1 2
Let's reduce by a power x 1 2, we get 4 x 1 2 - 1 · x 1 2 + 1 .
Additionally, you can simplify the power expression in the denominator using the difference of squares formula: squares: 4 x 1 2 - 1 x 1 2 + 1 = 4 x 1 2 2 - 1 2 = 4 x - 1 .
Answer: x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = 4 x - 1
Example 11
Simplify the power-law expression x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3.
Solution
We can reduce the fraction by (x 2 , 7 + 1) 2. We get the fraction x 3 4 x - 5 8 x 2, 7 + 1.
Let's continue transforming the powers of x x 3 4 x - 5 8 · 1 x 2, 7 + 1. Now you can use the property of dividing powers with the same bases: x 3 4 x - 5 8 1 x 2, 7 + 1 = x 3 4 - - 5 8 1 x 2, 7 + 1 = x 1 1 8 1 x 2 , 7 + 1 .
We move from the last product to the fraction x 1 3 8 x 2, 7 + 1.
Answer: x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3 = x 1 3 8 x 2, 7 + 1.
In most cases, it is more convenient to transfer factors with negative exponents from the numerator to the denominator and back, changing the sign of the exponent. This action allows you to simplify the further decision. Let's give an example: the power expression (x + 1) - 0, 2 3 · x - 1 can be replaced by x 3 · (x + 1) 0, 2.
Converting expressions with roots and powers
In problems there are power expressions that contain not only powers with fractional exponents, but also roots. It is advisable to reduce such expressions only to roots or only to powers. Going for degrees is preferable as they are easier to work with. This transition is especially preferable when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to access the modulus or split the ODZ into several intervals.
Example 12
Express the expression x 1 9 · x · x 3 6 as a power.
Solution
Range of permissible variable values x is defined by two inequalities x ≥ 0 and x x 3 ≥ 0, which define the set [ 0 , + ∞) .
On this set we have the right to move from roots to powers:
x 1 9 · x · x 3 6 = x 1 9 · x · x 1 3 1 6
Using the properties of powers, we simplify the resulting power expression.
x 1 9 · x · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 · 1 3 · 6 = = x 1 9 · x 1 6 x 1 18 = x 1 9 + 1 6 + 1 18 = x 1 3
Answer: x 1 9 · x · x 3 6 = x 1 3 .
Converting powers with variables in the exponent
These transformations are quite easy to make if you use the properties of the degree correctly. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.
We can replace by the product of powers, the exponents of which are the sum of some variable and a number. On the left side, this can be done with the first and last terms of the left side of the expression:
5 2 x 5 1 − 3 5 x 7 x − 14 7 2 x 7 − 1 = 0, 5 5 2 x − 3 5 x 7 x − 2 7 2 x = 0 .
Now let's divide both sides of the equation by 7 2 x. This expression for the variable x takes only positive values:
5 5 - 3 5 x 7 x - 2 7 2 x 7 2 x = 0 7 2 x , 5 5 2 x 7 2 x - 3 5 x 7 x 7 2 x - 2 7 2 x 7 2 x = 0 , 5 5 2 x 7 2 x - 3 5 x 7 x 7 x 7 x - 2 7 2 x 7 2 x = 0
Let's reduce fractions with powers, we get: 5 · 5 2 · x 7 2 · x - 3 · 5 x 7 x - 2 = 0.
Finally, the ratio of powers with the same exponents is replaced by powers of ratios, resulting in the equation 5 5 7 2 x - 3 5 7 x - 2 = 0, which is equivalent to 5 5 7 x 2 - 3 5 7 x - 2 = 0 .
Let us introduce a new variable t = 5 7 x, which reduces the solution of the original exponential equation to the solution of the quadratic equation 5 · t 2 − 3 · t − 2 = 0.
Converting expressions with powers and logarithms
Expressions containing powers and logarithms are also found in problems. An example of such expressions is: 1 4 1 - 5 · log 2 3 or log 3 27 9 + 5 (1 - log 3 5) · log 5 3. The transformation of such expressions is carried out using the approaches and properties of logarithms discussed above, which we discussed in detail in the topic “Transformation of logarithmic expressions”.
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Municipal state educational institution
basic secondary school No. 25
Algebra lesson
Subject:
« Converting expressions containing powers with fractional exponents"
Developed by:
,
mathematic teacher
higher toqualification category
Nodal
2013
Lesson topic: Converting expressions containing exponents with fractional exponents
The purpose of the lesson:
1. Further development of skills, knowledge, and skills in converting expressions containing degrees with fractional exponents
2. Development of the ability to find errors, development of thinking, creativity, speech, computing skills
3. Fostering independence, interest in the subject, attentiveness, accuracy.
TCO: magnetic board, test cards, tables, individual cards, schoolchildren have blank signed sheets on the table for individual work, a crossword puzzle, tables for mathematical warm-up, a multimedia projector.
Lesson type: securing ZUN.
Lesson plan over time
1. Organizational aspects (2 min)
2. Checking homework (5 min)
3. Crossword puzzle (3 min)
4. Mathematical warm-up (5 min)
5. Solving frontal strengthening exercises (7 min)
6. Individual work (10 min)
7. Solution of repetition exercises (5 min)
8. Lesson summary (2 min)
9. Homework assignment (1 min)
During the classes
1) Checking homework in the form of peer review . Good students check the notebooks of weak children. And the weak guys check with the strong ones using a sample control card. Homework is given in two versions.
I option the task is not difficult
II option the task is difficult
As a result of the check, the guys highlight the mistakes with a simple pencil and give a rating. I finally check the work after the children hand in their notebooks after class. I ask the guys the results of their test and put grades for this type of work in my summary table.
2) To test theoretical material, a crossword puzzle is offered.
Vertically:
1. Property of multiplication used when multiplying a monomial by a polynomial?
2. The effect of exponents when raising a power to a power?
3. A degree with zero index?
4. A product consisting of identical factors?
Horizontally:
5. Root n – oh degree of a non-negative number?
6. The action of exponents when multiplying powers?
7. The effect of exponents in dividing powers?
8. The number of all identical factors?
3) Mathematical warm-up
a) perform the calculation and use the cipher to read the word hidden in the problem.
There is a table on the board in front of you. The table in column 1 contains examples that need to be calculated.
Key to the table
491/2 | ||
27-1/3 | ||
4*81/3 | ||
5*25-1/2 | ||
7*82/3 | ||
(49/144)1/2 | 7/12 | |
(27*64)1/3 |
7/12 |
And write the answer in the column II, and in column III put the letter corresponding to this answer.
Teacher: So, the encrypted word is “degree”. In the next task we work with the 2nd and 3rd degrees
b) Game “Make sure you don’t make a mistake”
Instead of dots, put a number
a) x=(x...)2; b) a3/2 = (a1/2)…; c) a=(a1/3)…; d) 5... = (51/4)2; e) 34/3=(34/9)…; e) 74/5 = (7...)2; g) x1/2=(x...)2; h) y1/2=(y...)2
Let's find the error:
А1/4 – 2а1/2 + 1 = (а1/
So, guys, what needed to be used to complete this task:
Property of degrees: when raising a degree to a power, the exponents are multiplied;
4) Now let’s get started with the front-end written work. , using the results of previous work. Open notebooks and write down the date and topic of the lesson.
№ 000
a) a – b = (a1/2)2 – (b1/2)2 = (a1/2 – b1/2)*(a1/2 + b1/2)
b) a – c = (a1/3)3 – (b1/3)3 = (a1/3 – c1/3)*(a2/3 + a1/3 b1/3 + b2/3)
No. 000 (a, c, d, e)
A ) m2 – 5 = m2 – (m1/2)2 = (m – 51/2)*(m+51/2)
c) a3 – 4 = (a3/2)2 – 22 = (a3/2 – 2)*(a3/2 +2)
d) x2/5 – y4/5 = (x1/5)2 – (y2/5)2 = (x1/5 – y2/5)*(x1/5 + y2/5)
e) 4 – a = 22 – (a1/2)2 = (2 – a1/2)*(2+a1/2)
No. 000 (a, d, f)
a) x3 – 2 = x3 – (21/3)3 = (x – 21/3)*(x2 + 21/3 x + 22/3)
d) a6/5 + 27 = (a2/5)3 + 33 = (a2/5 + 3)*(a4/3 – 3 a2/5 + 9)
f) 4 + y = (41/3)3 + (y1/3)3 = (41/3 + y1/3)*(42/3 + 41/3 y1/3 + y2/3)
Grade
5) Work on individual cards using four options on separate sheets
Tasks with varying degrees of difficulty are completed without any prompting from the teacher.
I check the work immediately and put grades in my table and on the guys’ papers.
No. 000 (a, c, d, h)
a) 4*31/2/(31/2 – 3) = 4*31/2 /31/2*(1 – 31/2) = 4 / (1 – 31/2)
c) x + x1/2 /2x = x1/2*(x1/2+1)/ 2*(x1/2)2 = (x1/2+1)/ 2x1/2
e) (a2/3 – b2/3)/(a1/3 +b1/3) = (a1/3)2 – (b1/3)2/(a1/3 +b1/3) = (a1/3 + b1/3)*(a1/3 – b1/3)/(a1/3 + b1/3) = a1/3 – b1/3
h) (x2/3 - x1/3 y1/3 + y2/3)/(x + y) = ((x1/3)2 – x1/3 y1/3 + (y1/3)2)/(( x1/3)3 +(y1/3)3) = ((x1/3)2 – x1/3 y1/3 +(y1/3)2)/(x1/3 +y1/3)*((x1 /3)2 – x1/3 y1/3 + (y1/3)2) = 1/ (x1/3 + y1/3)
7) Working on individual cards with varying degrees of complexity. In some exercises there are recommendations from the teacher, since the material is complicated and it is difficult for weak children to cope with the work
There are also four options available. The assessment takes place immediately. I put all the grades in a spreadsheet.
Problem No. from the collection
The teacher asks questions:
1. What should be found in the problem?
2. What do you need to know for this?
3. How to express the time of 1 pedestrian and 2 pedestrians?
4. Compare the times of pedestrians 1 and 2 according to the conditions of the problem and create an equation.
The solution of the problem:
Let x (km/h) be the speed of 1 pedestrian
X +1 (km/h) – speed 2 pedestrians
4/х (h) – pedestrian time
4/(x +1) (h) – time of the second pedestrian
According to the conditions of the problem 4/x >4/ (x +1) for 12 minutes
12 min = 12 /60 h = 1/5 h
Let's make an equation
X/4 – 4/ (x +1) = 1/5
NOZ: 5x(x +1) ≠ 0
5*4*(x+1) – 5*4x = x*(x+1)
20x + 20 – 20x – x2 – x = 0
X2 +x –20 = 0
D=1 – 4*(-20) = 81, 81>0, 2 k
x1 = (-1 -√81)/(-2) = 5 km/h – speed of 1 pedestrian
x2 = (-1 + √81)/(-2) = 4 – does not fit the meaning of the problem, since x>0
Answer: 5 km/h – speed of 2 pedestrians
9) Lesson summary: So, guys, today in the lesson we consolidated knowledge, skills, and skills of transforming expressions containing degrees, applied abbreviated multiplication formulas, moved the common factor out of brackets, and repeated the material covered. I point out the advantages and disadvantages.
Summarizing the lesson in a table.
Crossword | Mat. warm-up | Front. Job | Ind. work K-1 | Ind. work K-2 | ||||
10) I announce the grades. Homework assignment
Individual cards K – 1 and K – 2
I change B – 1 and B – 2; B – 3 and B – 4, since they are equivalent
Applications to the lesson.
1) Cards for homework
1. simplify
a) (x1/2 – y1/2)2 + 2x1/2 y1/2
b) (a3/2 + 5a1\2)2 – 10a2
2. present as a sum
a) a1/3 c1\4*(b2/3 + c3/4)
b) (a1/2 – b1/2)*(a + a1/2 b1\2 + c)
3. take out the overall multiplier
c) 151/3 +201/3
1. simplify
a) √m + √n – (m1/4 – n1/4)2
b) (a1/4 + b1/4)*(a1/8 + b1/8)*(a1\8 – b1/8)
2. present as a sum
a) x0.5 y0.5*(x-0.5 – y1.5)
b) (x1/3 + y1/3)*(x2\3 – x1/3 y1\3 + y2/3)
3. Take the common factor out of brackets
b) c1\3 – c
c) (2a)1/3 – (5a)1\3
2) control card for B – 2
a) √m + √n – (m 1|4 – n 1|4)2 = m 1|2 + n 1|2 – ((m 1|2)2 – 2 m 1/4 n 1/4 + (n 1/2)2) = m 1/2 + n 1/2 – m 1/2 + 2 m 1/4 n 1/4 – n 1/2 = 2 m 1/4 n 1/4
b) (a1/4 + b1/4)*(a1/8 + b1/8)*(a1/8 – b1/8) = (a1/4 + b1/4)*(a1/8)2 – ( в1/8)2 = (а1/4 + в1/4)*(а1/4 – в1/4) = (а1/4)2 – (в1/4)2 = а1/2 – в1/2
a) x0.5 y0.5* (x-0.5-y1.5) = x0.5 y0.5 x-0.5 – x0.5 y0.5y1.5 = x0 y0.5 – x0.5 y2 = y0.5 – x0.5 y2
b) (x1/3 + y1/3)*(x2/3 – x1/3 y1\3 + y2/3) = (x1\3 + y1/3)*((x1/3)2 – x1/3 y1\3 + (y1/3)2) = (x1/3)2 + (y1/3)2 = x + y
a) 3 – 31/2 = 31/2 * (31/2 - 1)
b) v1/3 – v = v1/3 *(1 – v2/3)
c) (2a)1/3 – (5a)1/3 = a1/3*(21/3 – 51/3)
3) Cards for the first individual work
a) a – y, x ≥ 0, y ≥ 0
b) a – and, a ≥ 0
1. Factorize as a difference of squares
a) a1/2 – b1/2
2. Factorize as a difference or sum of cubes
a) c1/3 + d1/3
1. Factorize as a difference of squares
a) X1/2 + Y1/2
b) X1/4 – U1/4
2. Factorize as a difference or sum of cubes
4) cards for the second individual work
a) (x – x1/2)/ (x1/2 – 1)
Instruction: x1/2, remove numerators from brackets
b) (a - c)/(a1/2 – b1/2)
Note: a – b = (a1/2)2 – (b1/2)2
Reduce the fraction
a) (21/4 – 2)/ 5*21/4
Instruction: remove 21/4 from brackets
b) (a – c)/(5а1/2 – 5в1/2)
Note: a – b = (a1/2)2 – (b1/2)2
Option 3
1. Reduce the fraction
a) (x1/2 – x1/4)/x3/4
Instruction: put x1/4 out of brackets
b) (a1/2 – b1/2)/(4a1/4 – 4b1/4)
Option 4
Reduce the fraction
a) 10/ (10 – 101/2)
b) (a - c)/(a2/3 + a1\3b1/3+ B 1/3)
An expression of the form a (m/n), where n is some natural number, m is some integer and the base of the degree a is greater than zero, called a degree with a fractional exponent. Moreover, the following equality is true. n√(a m) = a (m/n) .
As we already know, numbers of the form m/n, where n is some natural number and m is some integer, are called fractional or rational numbers. From all of the above we obtain that the degree is defined for any rational exponent and any positive base of the degree.
For any rational numbers p,q and any a>0 and b>0 the following equalities are true:
- 1. (a p)*(a q) = a (p+q)
- 2. (a p):(b q) = a (p-q)
- 3. (a p) q = a (p*q)
- 4. (a*b) p = (a p)*(b p)
- 5. (a/b) p = (a p)/(b p)
These properties are widely used when converting various expressions that contain powers with fractional exponents.
Examples of transformations of expressions containing powers with a fractional exponent
Let's look at a few examples that demonstrate how these properties can be used to transform expressions.
1. Calculate 7 (1/4) * 7 (3/4) .
- 7 (1/4) * 7 (3/4) = z (1/4 + 3/4) = 7.
2. Calculate 9 (2/3) : 9 (1/6) .
- 9 (2/3) : 9 (1/6) = 9 (2/3 - 1/6) = 9 (1/2) = √9 = 3.
3. Calculate (16 (1/3)) (9/4) .
- (16 (1/3)) (9/4) = 16 ((1/3)*(9/4)) =16 (3/4) = (2 4) (3/4) = 2 (4*3/4) = 2 3 = 8.
4. Calculate 24 (2/3) .
- 24 (2/3) = ((2 3)*3) (2/3) = (2 (2*2/3))*3 (2/3) = 4*3√(3 2)=4*3√9.
5. Calculate (8/27) (1/3) .
- (8/27) (1/3) = (8 (1/3))/(27 (1/3)) = ((2 3) (1/3))/((3 3) (1/3))= 2/3.
6. Simplify the expression ((a (4/3))*b + a*b (4/3))/(3√a + 3√b)
- ((a (4/3))*b + a*b (4/3))/(3√a + 3√b) = (a*b*(a (1/3) + b (1/3 )))/(1/3) + b (1/3)) = a*b.
7. Calculate (25 (1/5))*(125 (1/5)).
- (25 (1/5))*(125 (1/5)) =(25*125) (1/5) = (5 5) (1/5) = 5.
8. Simplify the expression
- (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a (2/3) + a (-1/3)).
- (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a (2/3) + a (-1/3)) =
- = ((a (1/3))*(1-a 2))/((a (1/3))*(1-a)) - ((a (-1/3))*(1- a 2))/ ((a (-1/3))*(1+a)) =
- = 1 +a - (1-a) = 2*a.
As you can see, using these properties, you can significantly simplify some expressions that contain powers with fractional exponents.
Expressions, expression conversion
Power expressions (expressions with powers) and their transformation
In this article we will talk about converting expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening parentheses and bringing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.
Page navigation.
What are power expressions?
The term “power expressions” practically does not appear in school mathematics textbooks, but it appears quite often in collections of problems, especially those intended for preparation for the Unified State Exam and the Unified State Exam, for example. After analyzing the tasks in which it is necessary to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing powers in their entries. Therefore, you can accept the following definition for yourself:
Definition.
Power expressions are expressions containing degrees.
Let's give examples of power expressions. Moreover, we will present them according to how the development of views on from a degree with a natural exponent to a degree with a real exponent occurs.
As is known, first one gets acquainted with the power of a number with a natural exponent; at this stage, the first simplest power expressions of the type 3 2, 7 5 +1, (2+1) 5, (−0.1) 4, 3 a 2 appear −a+a 2 , x 3−1 , (a 2) 3 etc.
A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with negative integer powers, like the following: 3 −2, 
, a −2 +2 b −3 +c 2 .
In high school they return to degrees. There, a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: 
, , 
and so on. Finally, degrees with irrational exponents and expressions containing them are considered: , .
The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and, for example, the following expressions arise: 2 x 2 +1 or 
. And after getting acquainted with , expressions with powers and logarithms begin to appear, for example, x 2·lgx −5·x lgx.
So, we have dealt with the question of what power expressions represent. Next we will learn to transform them.
Main types of transformations of power expressions
With power expressions, you can perform any of the basic identity transformations of expressions. For example, you can open parentheses, replace numerical expressions with their values, add similar terms, etc. Naturally, in this case, it is necessary to follow the accepted procedure for performing actions. Let's give examples.
Example.
Calculate the value of the power expression 2 3 ·(4 2 −12) .
Solution.
According to the order of execution of actions, first perform the actions in brackets. There, firstly, we replace the power 4 2 with its value 16 (if necessary, see), and secondly, we calculate the difference 16−12=4. We have 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4.
In the resulting expression, we replace the power 2 3 with its value 8, after which we calculate the product 8·4=32. This is the desired value.
So, 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4=8·4=32.
Answer:
2 3 ·(4 2 −12)=32.
Example.
Simplify expressions with powers 3 a 4 b −7 −1+2 a 4 b −7.
Solution.
Obviously, this expression contains similar terms 3·a 4 ·b −7 and 2·a 4 ·b −7 , and we can present them: .
Answer:
3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.
Example.
Express an expression with powers as a product.
Solution.
You can cope with the task by representing the number 9 as a power of 3 2 and then using the formula for abbreviated multiplication - difference of squares: 
Answer:
There are also a number of identical transformations inherent specifically in power expressions. We will analyze them further.
Working with base and exponent
There are powers whose base and/or exponent are not just numbers or variables, but some expressions. As an example, we give the entries (2+0.3·7) 5−3.7 and (a·(a+1)−a 2) 2·(x+1) .
When working with such expressions, you can replace both the expression in the base of the degree and the expression in the exponent with an identically equal expression in the ODZ of its variables. In other words, according to the rules known to us, we can separately transform the base of the degree and separately the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.
Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the power expression mentioned above (2+0.3 7) 5−3.7, you can perform operations with the numbers in the base and exponent, which will allow you to move to the power 4.1 1.3. And after opening the brackets and bringing similar terms to the base of the degree (a·(a+1)−a 2) 2·(x+1), we obtain a power expression of a simpler form a 2·(x+1) .
Using Degree Properties
One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following properties of powers are true:
- a r ·a s =a r+s ;
- a r:a s =a r−s ;
- (a·b) r =a r ·b r ;
- (a:b) r =a r:b r ;
- (a r) s =a r·s .
Note that for natural, integer, and positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n the equality a m ·a n =a m+n is true not only for positive a, but also for negative a, and for a=0.
At school, the main focus when transforming power expressions is on the ability to choose the appropriate property and apply it correctly. In this case, the bases of degrees are usually positive, which allows the properties of degrees to be used without restrictions. The same applies to the transformation of expressions containing variables in the bases of powers - the range of permissible values of variables is usually such that the bases take only positive values on it, which allows you to freely use the properties of powers. In general, you need to constantly ask yourself whether it is possible to use any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the educational value and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using properties of degrees. Here we will limit ourselves to considering a few simple examples.
Example.
Express the expression a 2.5 ·(a 2) −3:a −5.5 as a power with base a.
Solution.
First, we transform the second factor (a 2) −3 using the property of raising a power to a power: (a 2) −3 =a 2·(−3) =a −6. The original power expression will take the form a 2.5 ·a −6:a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 ·a −6:a −5.5 =
a 2.5−6:a −5.5 =a −3.5:a −5.5 =
a −3.5−(−5.5) =a 2 .
Answer:
a 2.5 ·(a 2) −3:a −5.5 =a 2.
Properties of powers when transforming power expressions are used both from left to right and from right to left.
Example.
Find the value of the power expression.
Solution.
The equality (a·b) r =a r ·b r, applied from right to left, allows us to move from the original expression to a product of the form and further. And when multiplying powers with the same bases, the exponents add up: 
.
It was possible to transform the original expression in another way: 
Answer:
.
Example.
Given the power expression a 1.5 −a 0.5 −6, introduce a new variable t=a 0.5.
Solution.
The degree a 1.5 can be represented as a 0.5 3 and then, based on the property of the degree to the degree (a r) s =a r s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6=(a 0.5) 3 −a 0.5 −6. Now it’s easy to introduce a new variable t=a 0.5, we get t 3 −t−6.
Answer:
t 3 −t−6 .
Converting fractions containing powers
Power expressions can contain or represent fractions with powers. Any of the basic transformations of fractions that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain powers can be reduced, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate these words, consider solutions to several examples.
Example.
Simplify power expression 
.
Solution.
This power expression is a fraction. Let's work with its numerator and denominator. In the numerator we open the brackets and simplify the resulting expression using the properties of powers, and in the denominator we present similar terms: 
And let’s also change the sign of the denominator by placing a minus in front of the fraction: 
.
Answer:
.
Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the ODZ. To prevent this from happening, it is necessary that the additional factor does not go to zero for any values of the variables from the ODZ variables for the original expression.
Example.
Reduce the fractions to a new denominator: a) to denominator a, b) 
to the denominator.
Solution.
a) In this case, it is quite easy to figure out which additional multiplier helps to achieve the desired result. This is a multiplier of a 0.3, since a 0.7 ·a 0.3 =a 0.7+0.3 =a. Note that in the range of permissible values of the variable a (this is the set of all positive real numbers), the power of a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of a given fraction by this additional factor: 
b) Taking a closer look at the denominator, you will find that 
and multiplying this expression by will give the sum of cubes and , that is, . And this is the new denominator to which we need to reduce the original fraction.
This is how we found the additional factor. In the range of permissible values of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it: 
Answer:
A) 
, b) 
.
There is also nothing new in reducing fractions containing powers: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are reduced.
Example.
Reduce the fraction: a) 
, b) .
Solution.
a) Firstly, the numerator and denominator can be reduced by the numbers 30 and 45, which is equal to 15. It is also obviously possible to perform a reduction by x 0.5 +1 and by 
. Here's what we have: 
b) In this case, identical factors in the numerator and denominator are not immediately visible. To obtain them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator using the difference of squares formula: 
Answer:
A) 
b) 
.
Converting fractions to a new denominator and reducing fractions are mainly used to do things with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), but the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its inverse.
Example.
Follow the steps 
.
Solution.
First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is 
, after which we subtract the numerators: 
Now we multiply the fractions: 
Obviously, it is possible to reduce by a power of x 1/2, after which we have 
.
You can also simplify the power expression in the denominator by using the difference of squares formula: 
.
Answer:
Example.
Simplify the Power Expression 
.
Solution.
Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction 
. It is clear that something else needs to be done with the powers of X. To do this, we transform the resulting fraction into a product. This gives us the opportunity to take advantage of the property of dividing powers with the same bases: 
. And at the end of the process we move from the last product to the fraction.
Answer:
.
And let us also add that it is possible, and in many cases desirable, to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .
Converting expressions with roots and powers
Often, in expressions in which some transformations are required, roots with fractional exponents are also present along with powers. To transform such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with powers, they usually move from roots to powers. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODZ into several intervals (we discussed this in detail in the article transition from roots to powers and back After getting acquainted with the degree with a rational exponent a degree with an irrational exponent is introduced, which allows us to talk about a degree with an arbitrary real exponent. At this stage, it begins to be studied at school. exponential function, which is analytically given by a power, the base of which is a number, and the exponent is a variable. So we are faced with power expressions containing numbers in the base of the power, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.
It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations And exponential inequalities, and these conversions are quite simple. In the overwhelming majority of cases, they are based on the properties of the degree and are aimed, for the most part, at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.
Firstly, powers, in the exponents of which is the sum of a certain variable (or expression with variables) and a number, are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 =0,
5 5 2 x −3 5 x 7 x −2 7 2 x =0.
Next, both sides of the equality are divided by the expression 7 2 x, which on the ODZ of the variable x for the original equation takes only positive values (this is a standard technique for solving equations of this type, we are not talking about it now, so focus on subsequent transformations of expressions with powers ): 
Now we can cancel fractions with powers, which gives 
.
Finally, the ratio of powers with the same exponents is replaced by powers of relations, resulting in the equation 
, which is equivalent 
. The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of a quadratic equation
