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How to find the value of an expression with fractional powers. Equations online

Let's consider the topic of transforming expressions with powers, but first we will dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open brackets, give like terms, work with the base and exponent, use the properties of powers.

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What are Power Expressions?

In the school course, few people use the phrase "power expressions", but this term is constantly found in collections for preparing for the exam. In most cases, the 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.

We give several examples of power expressions, starting with a degree with a natural exponent and ending with a degree 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 . As well as 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 a variable 3 x - 54 - 7 3 x - 58 or a 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 transform them.

The main types of transformations of power expressions

First of all, we will consider the basic identity transformations of expressions that can be performed with power expressions.

Example 1

Calculate Power Expression Value 2 3 (4 2 − 12).

Solution

We will carry out all transformations in compliance with the order of actions. IN this case We'll start by doing the parentheses: we'll replace the exponent with a numeric value and calculate the difference between the two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.

It remains for us to replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here is our answer.

Answer: 2 3 (4 2 − 12) = 32 .

Example 2

Simplify expression with powers 3 a 4 b − 7 − 1 + 2 a 4 b − 7.

Solution

The expression given to us in the condition of the problem contains similar terms, which we can bring: 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 an expression with powers of 9 - b 3 · π - 1 2 as a product.

Solution

Let's represent 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 .

And now let's move on to the analysis of identical 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 . It is difficult to work with such records. 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.

The transformations of the degree and the indicator are carried out according to the rules known to us separately from each other. The most important thing is that as a result of the transformations, an expression is obtained that is identical to the original one.

The purpose of transformations is to simplify the original expression or to obtain a solution to the problem. For example, in the example we gave above, (2 + 0 , 3 7) 5 − 3 , 7 you can perform operations to go to the degree 4 , 1 1 , 3 . Opening the brackets, we can bring like terms in the base of the degree (a (a + 1) − a 2) 2 (x + 1) and get a power expression over simple form a 2 (x + 1).

Using Power Properties

The properties of degrees, written as equalities, are one of the main tools for transforming expressions with degrees. We present here the main ones, considering 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 stringent. 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.

You can apply the properties of degrees without restrictions in cases where the bases of the degrees are positive or contain variables, the area allowed values which is such that on it the grounds accept only positive values. In fact, within the framework of the school curriculum in mathematics, the task of the student is to choose the appropriate property and apply it correctly.

When preparing for admission to universities, there may be tasks in which inaccurate application of properties will lead to a narrowing of the ODZ and other difficulties with the solution. In this section, we will consider only two such cases. More information on the subject can be found in the topic "Transforming expressions using exponent properties".

Example 4

Represent the expression a 2 , 5 (a 2) - 3: a - 5 , 5 as a degree with a base a.

Solution

To begin with, we use the exponentiation property 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 .

The transformation of power expressions according to the property of degrees 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 the equality (a b) r = a r b r, from right to left, then 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 \u003d 21 1 3 + 2 3 \u003d 21 1 \u003d 21.

There is another way to make transformations:

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

Imagine the degree a 1 , 5 How a 0 , 5 3. Using the degree property in a degree (a r) s = a r s from right to left and get (a 0 , 5) 3: a 1 , 5 - a 0 , 5 - 6 = (a 0 , 5) 3 - a 0 , 5 - 6 . In the resulting expression, you can easily introduce a new variable t = a 0 , 5: get t 3 − t − 6.

Answer: t 3 − t − 6 .

Converting fractions containing powers

We usually deal with two variants of power expressions with fractions: the expression is a fraction with a degree or contains such a fraction. All basic fraction transformations are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, work 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

Put a minus 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 vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example 8

Bring 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) We choose 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 take a 0 , 3. The range of admissible values ​​of the variable a includes the set of all positive real numbers. In this area, the degree 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) 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

Multiply this expression by x 1 3 + 2 · y 1 6 , we get the sum of 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 bring the original fraction.

So we found an additional factor x 1 3 + 2 · y 1 6 . On the range of acceptable values ​​of variables x And y the expression x 1 3 + 2 y 1 6 does not vanish, so 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) Use the greatest common denominator (GCD) by which the numerator and denominator can be reduced. For the numbers 30 and 45, this is 15 . We can also reduce x 0 , 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 .

The basic operations with fractions include reduction to a new denominator and reduction of fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, the fractions are first reduced to common denominator, after which operations (addition or subtraction) are performed with 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 brackets. 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 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 degree 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 formula for the difference of squares: 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 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 a fraction x 3 4 x - 5 8 x 2, 7 + 1.

Let's continue transformations of x powers x 3 4 x - 5 8 · 1 x 2 , 7 + 1 . Now you can use the power division property 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 pass 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 multipliers with negative exponents from the numerator to the denominator and vice versa by changing the sign of the exponent. This action simplifies 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 tasks, there are power expressions that contain not only degrees with fractional exponents, but also roots. It is desirable to reduce such expressions only to roots or only to powers. The transition to degrees is preferable, since they are easier to work with. Such a transition is especially advantageous when the DPV of the variables for the original expression allows you to replace the roots with powers without having to access the modulus or split the DPV into several intervals.

Example 12

Express the expression x 1 9 x x 3 6 as a power.

Solution

Valid range of a variable x is determined 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 degrees, 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 simple to make if you correctly use the properties of the degree. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.

We can replace the product of the degree, in terms of which the sum of some variable and a number is found. On the left side, this can be done with the first and last terms on 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 on the ODZ of 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 the 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, which leads to 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 a decision 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. Examples of such expressions are: 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 discussed above and the properties of logarithms, which we have analyzed in detail in the topic “Transformation of logarithmic expressions”.

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

The exponent is used to make it easier to write the operation of multiplying a number by itself. For example, instead of writing, you can write 4 5 (\displaystyle 4^(5))(an explanation of such a transition is given in the first section of this article). Powers make it easier to write long or complex expressions or equations; also, powers are easily added and subtracted, resulting in a simplification of an expression or equation (for example, 4 2 ∗ 4 3 = 4 5 (\displaystyle 4^(2)*4^(3)=4^(5))).

Note: if you need to solve an exponential equation (in such an equation, the unknown is in the exponent), read.

Steps

Solving simple problems with powers

Multiply the base of the exponent by itself a number of times equal to the exponent. If you need to solve a problem with exponents manually, rewrite the exponent as a multiplication operation, where the base of the exponent is multiplied by itself. For example, given the degree 3 4 (\displaystyle 3^(4)). In this case, the base of degree 3 must be multiplied by itself 4 times: 3 ∗ 3 ∗ 3 ∗ 3 (\displaystyle 3*3*3*3). Here are other examples:

First, multiply the first two numbers. For example, 4 5 (\displaystyle 4^(5)) = 4 ∗ 4 ∗ 4 ∗ 4 ∗ 4 (\displaystyle 4*4*4*4*4). Don't worry - the calculation process is not as complicated as it seems at first glance. First multiply the first two quadruples, and then replace them with the result. Like this:

• 4 5 = 4 ∗ 4 ∗ 4 ∗ 4 ∗ 4 (\displaystyle 4^(5)=4*4*4*4*4)
  • 4 ∗ 4 = 16 (\displaystyle 4*4=16)

1. Multiply the result (16 in our example) by the next number. Each subsequent result will increase proportionally. In our example, multiply 16 by 4. Like this:

   • 4 5 = 16 ∗ 4 ∗ 4 ∗ 4 (\displaystyle 4^(5)=16*4*4*4)
     • 16 ∗ 4 = 64 (\displaystyle 16*4=64)
   • 4 5 = 64 ∗ 4 ∗ 4 (\displaystyle 4^(5)=64*4*4)
     • 64 ∗ 4 = 256 (\displaystyle 64*4=256)
   • 4 5 = 256 ∗ 4 (\displaystyle 4^(5)=256*4)
     • 256 ∗ 4 = 1024 (\displaystyle 256*4=1024)
   • Keep multiplying the result of multiplying the first two numbers by the next number until you get the final answer. To do this, multiply the first two numbers, and then multiply the result by the next number in the sequence. This method is valid for any degree. In our example, you should get: 4 5 = 4 ∗ 4 ∗ 4 ∗ 4 ∗ 4 = 1024 (\displaystyle 4^(5)=4*4*4*4*4=1024) .

2. Solve the following problems. Check your answer with a calculator.

   • 8 2 (\displaystyle 8^(2))
   • 3 4 (\displaystyle 3^(4))
   • 10 7 (\displaystyle 10^(7))

3. On the calculator, look for the key labeled "exp", or " x n (\displaystyle x^(n))", or "^". With this key you will raise a number to a power. It is practically impossible to manually calculate the degree with a large exponent (for example, the degree 9 15 (\displaystyle 9^(15))), but the calculator can easily cope with this task. In Windows 7, the standard calculator can be switched to engineering mode; to do this, click "View" -\u003e "Engineering". To switch to normal mode, click "View" -\u003e "Normal".

   • Check your answer with search engine(Google or Yandex). Using the "^" key on the computer keyboard, enter the expression into the search engine, which will instantly display the correct answer (and possibly suggest similar expressions for study).

   Addition, subtraction, multiplication of powers

   1. You can add and subtract powers only if they have the same base. If you need to add powers with the same bases and exponents, then you can replace the addition operation with a multiplication operation. For example, given the expression 4 5 + 4 5 (\displaystyle 4^(5)+4^(5)). Remember that the degree 4 5 (\displaystyle 4^(5)) can be represented as 1 ∗ 4 5 (\displaystyle 1*4^(5)); Thus, 4 5 + 4 5 = 1 ∗ 4 5 + 1 ∗ 4 5 = 2 ∗ 4 5 (\displaystyle 4^(5)+4^(5)=1*4^(5)+1*4^(5) =2*4^(5))(where 1 +1 =2). That is, count the number of similar degrees, and then multiply such a degree and this number. In our example, raise 4 to the fifth power, and then multiply the result by 2. Remember that the addition operation can be replaced by a multiplication operation, for example, 3 + 3 = 2 ∗ 3 (\displaystyle 3+3=2*3). Here are other examples:

      • 3 2 + 3 2 = 2 ∗ 3 2 (\displaystyle 3^(2)+3^(2)=2*3^(2))
      • 4 5 + 4 5 + 4 5 = 3 ∗ 4 5 (\displaystyle 4^(5)+4^(5)+4^(5)=3*4^(5))
      • 4 5 − 4 5 + 2 = 2 (\displaystyle 4^(5)-4^(5)+2=2)
      • 4 x 2 − 2 x 2 = 2 x 2 (\displaystyle 4x^(2)-2x^(2)=2x^(2))

   2. When multiplying powers with the same base, their exponents are added (the base does not change). For example, given the expression x 2 ∗ x 5 (\displaystyle x^(2)*x^(5)). In this case, you just need to add the indicators, leaving the base unchanged. Thus, x 2 ∗ x 5 = x 7 (\displaystyle x^(2)*x^(5)=x^(7)). Here is a visual explanation of this rule:

      When raising a power to a power, the exponents are multiplied. For example, given a degree. Since the exponents are multiplied, then (x 2) 5 = x 2 ∗ 5 = x 10 (\displaystyle (x^(2))^(5)=x^(2*5)=x^(10)). The meaning of this rule is that you multiply the power (x 2) (\displaystyle (x^(2))) on itself five times. Like this:

      • (x 2) 5 (\displaystyle (x^(2))^(5))
      • (x 2) 5 = x 2 ∗ x 2 ∗ x 2 ∗ x 2 ∗ x 2 (\displaystyle (x^(2))^(5)=x^(2)*x^(2)*x^( 2)*x^(2)*x^(2))
      • Since the base is the same, the exponents simply add up: (x 2) 5 = x 2 ∗ x 2 ∗ x 2 ∗ x 2 ∗ x 2 = x 10 (\displaystyle (x^(2))^(5)=x^(2)*x^(2)* x^(2)*x^(2)*x^(2)=x^(10))

   3. An exponent with a negative exponent should be converted to a fraction (to the inverse power). It doesn't matter if you don't know what a reciprocal is. If you are given a degree with a negative exponent, for example, 3 − 2 (\displaystyle 3^(-2)), write this power in the denominator of the fraction (put 1 in the numerator), and make the exponent positive. In our example: 1 3 2 (\displaystyle (\frac (1)(3^(2)))). Here are other examples:

      When dividing powers with the same base, their exponents are subtracted (the base does not change). The division operation is the opposite of the multiplication operation. For example, given the expression 4 4 4 2 (\displaystyle (\frac (4^(4))(4^(2)))). Subtract the exponent in the denominator from the exponent in the numerator (do not change the base). Thus, 4 4 4 2 = 4 4 − 2 = 4 2 (\displaystyle (\frac (4^(4))(4^(2)))=4^(4-2)=4^(2)) = 16 .

      • The degree in the denominator can be written as follows: 1 4 2 (\displaystyle (\frac (1)(4^(2)))) = 4 − 2 (\displaystyle 4^(-2)). Remember that a fraction is a number (power, expression) with a negative exponent.

   4. Below are some expressions to help you learn how to solve power problems. The above expressions cover the material presented in this section. To see the answer, just highlight the empty space after the equals sign.

      Solving problems with fractional exponents

      1. A degree with a fractional exponent (for example, ) is converted to a root extraction operation. In our example: x 1 2 (\displaystyle x^(\frac (1)(2))) = x(\displaystyle(\sqrt(x))). It does not matter what number is in the denominator of the fractional exponent. For example, x 1 4 (\displaystyle x^(\frac (1)(4))) is the fourth root of "x" x 4 (\displaystyle (\sqrt[(4)](x))) .

      2. If the exponent is an improper fraction, then such an exponent can be decomposed into two powers to simplify the solution of the problem. There is nothing complicated about this - just remember the rule for multiplying powers. For example, given a degree. Turn that exponent into a root whose exponent is equal to the denominator of the fractional exponent, and then raise that root to the exponent equal to the numerator of the fractional exponent. To do this, remember that 5 3 (\displaystyle (\frac (5)(3))) = (1 3) ∗ 5 (\displaystyle ((\frac (1)(3)))*5). In our example:

         • x 5 3 (\displaystyle x^(\frac (5)(3)))
         • x 1 3 = x 3 (\displaystyle x^(\frac (1)(3))=(\sqrt[(3)](x)))
         • x 5 3 = x 5 ∗ x 1 3 (\displaystyle x^(\frac (5)(3))=x^(5)*x^(\frac (1)(3))) = (x 3) 5 (\displaystyle ((\sqrt[(3)](x)))^(5))
      3. Some calculators have a button for calculating exponents (first you need to enter the base, then press the button, and then enter the exponent). It is denoted as ^ or x^y.
      4. Remember that any number is equal to itself to the first power, for example, 4 1 = 4. (\displaystyle 4^(1)=4.) Moreover, any number multiplied or divided by one is equal to itself, for example, 5 ∗ 1 = 5 (\displaystyle 5*1=5) And 5 / 1 = 5 (\displaystyle 5/1=5).
      5. Know that the degree 0 0 does not exist (such a degree has no solution). When you try to solve such a degree on a calculator or on a computer, you will get an error. But remember that any number to the power of zero is equal to 1, for example, 4 0 = 1. (\displaystyle 4^(0)=1.)
      6. IN higher mathematics, which operates on imaginary numbers: e a i x = c o s a x + i s i n a x (\displaystyle e^(a)ix=cosax+isinax), Where i = (− 1) (\displaystyle i=(\sqrt (())-1)); e is a constant approximately equal to 2.7; a is an arbitrary constant. The proof of this equality can be found in any textbook on higher mathematics.

      Warnings

      • As the exponent increases, its value greatly increases. Therefore, if the answer seems wrong to you, in fact it may turn out to be true. You can check this by plotting any exponential function, such as 2 x .

Expressions, expression conversion

Power expressions (expressions with powers) and their transformation

In this article, we will talk about transforming expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening brackets, reducing 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.

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What are Power Expressions?

The term "power expressions" is practically not found in school textbooks of mathematics, but it often appears in collections of problems, especially designed to prepare for the Unified State Examination and the OGE, for example,. After analyzing tasks in which it is required to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing degrees in their entries. Therefore, for yourself, you can take the following definition:

Definition.

Power expressions are expressions containing powers.

Let's bring examples of power expressions. Moreover, we will represent them according to how the development of views on from a degree with a natural indicator to a degree with a real indicator takes place.

As you know, first you get acquainted with the degree 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 −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 the senior classes, they return to the degrees again. There is introduced a degree with rational indicator, which leads to 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 there are, for example, such expressions 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 figured out the question of what are power expressions. Next, we will learn how to transform them.

The 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 expand brackets, replace numeric expressions with their values, add like terms, and so on. 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 the actions, we first perform the actions in brackets. There, firstly, we replace the power of 4 2 with its value 16 (see if necessary), 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 of 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 Power Expressions 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 reduce 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.

To cope with the task allows the representation of the number 9 as a power of 3 2 and the subsequent use of the abbreviated multiplication formula, the difference of squares:

Answer:

There are also a number of identical transformations inherent in power expressions. Next, we will analyze them.

Working with base and exponent

There are degrees, in the basis and / or indicator of which are not just numbers or variables, but some expressions. As an example, let's write (2+0.3 7) 5−3.7 and (a (a+1)−a 2) 2 (x+1) .

When working with such expressions, it is possible to replace both the expression in the base of the degree and the expression in the indicator with an identically equal expression on the DPV of its variables. In other words, according to the rules known to us, we can separately convert the base of the degree, and separately - the indicator. It is clear that as a result of this transformation, an expression is 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 (2+0.3 7) 5−3.7 mentioned above, you can perform operations with numbers in the base and exponent, which will allow you to go to the power of 4.1 1.3. And after opening the brackets and bringing like terms in the base of the degree (a·(a+1)−a 2) 2·(x+1) we get a power expression of a simpler form a 2·(x+1) .

Using Power 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 power properties hold:

• 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, 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 ones, and for a=0 .

At school, the main attention in the transformation of power expressions is focused precisely on the ability to choose suitable property and apply it correctly. In this case, the bases of the degrees are usually positive, which allows you to use the properties of the degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of acceptable values ​​​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of degrees. In general, you need to constantly ask yourself whether it is possible to apply any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the ODZ and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using the properties of degrees. Here we confine ourselves to 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 by the property of raising a power to a power: (a 2) −3 =a 2 (−3) =a −6. In this case, the initial 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 \u003d a 2.

Power properties are used when transforming power expressions both from left to right and from right to left.

Example.

Find the value of the power expression.

Solution.

Equality (a·b) r =a r ·b r , applied from right to left, allows you to go from the original expression to the product of the form and further. And when multiplying powers with the same base, the indicators add up: .

It was possible to perform the transformation of the original expression in another way:

Answer:

.

Example.

Given a power expression a 1.5 −a 0.5 −6 , enter a new variable t=a 0.5 .

Solution.

The degree a 1.5 can be represented as a 0.5 3 and further on the basis of the property of the degree in the degree (a r) s =a r s applied from right to left, convert 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 is 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 fractions with powers or represent such fractions. Any of the basic fraction transformations that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain degrees can be reduced, reduced to a new denominator, work separately with their numerator and separately with the denominator, etc. To illustrate the above words, consider the solutions of 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 expression obtained after that using the properties of powers, and in the denominator we present similar terms:

And we also change the sign of the denominator by placing a minus in front of the fraction: .

Answer:

.

Reduction of containing powers of fractions to a new denominator is carried out similarly to reduction to a new denominator rational fractions. At the same time, 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 DPV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example.

Bring the fractions to a new denominator: a) to the denominator a, b) to the denominator.

Solution.

a) In this case, it is quite easy to figure out what additional factor helps to achieve the desired result. This is a factor a 0.3 since a 0.7 a 0.3 = a 0.7+0.3 = a . Note that on the range of acceptable values ​​of the variable a (this is the set of all positive real numbers), the degree a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of the given fraction by this additional factor:

b) Looking more closely at the denominator, we 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 bring the original fraction.

So we found an additional factor . The expression does not vanish on the range of acceptable values ​​of the variables x and y, therefore, we can multiply the numerator and denominator of the fraction by it:

Answer:

A) , b) .

There is also nothing new in the reduction of fractions containing degrees: the numerator and denominator are represented as a certain number of factors, and the same factors of the numerator and denominator are reduced.

Example.

Reduce the fraction: a) , b).

Solution.

a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which equals 15. Also, obviously, you can reduce by x 0.5 +1 and by . Here's what we have:

b) In this case, the same factors in the numerator and denominator are not immediately visible. To get them, you have to perform preliminary transformations. In this case, they consist in decomposing the denominator into factors according to the difference of squares formula:

Answer:

A)

b) .

Reducing fractions to a new denominator and reducing fractions is mainly used to perform operations on 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), and 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 reciprocal.

Example.

Follow the steps .

Solution.

First, we subtract the fractions in brackets. To do this, we bring them to a common denominator, which is , then subtract the numerators:

Now we multiply fractions:

Obviously, a reduction by the power x 1/2 is possible, after which we have .

You can also simplify the power expression in the denominator by using the difference of squares formula: .

Answer:

Example.

Simplify 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 convert the resulting fraction into a product. This gives us the opportunity to use the property of dividing powers with the same bases: . And at the end of the process, we pass from the last product to the fraction.

Answer:

.

And we 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 by 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, along with degrees with fractional exponents, there are also roots. To convert such an expression to the right kind, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with degrees, they usually move from roots to degrees. 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 degrees without the need to access the module or split the ODZ into several intervals (we discussed this in detail in the article, the transition from roots to powers and vice versa After getting acquainted with the degree with a rational exponent a degree with an irrational indicator is introduced, which makes it possible to speak of a degree with an arbitrary real indicator.At this stage, the school begins to study exponential function , which is analytically given by the degree, in the basis of which there is a number, and in the indicator - a variable. So we are faced with power expressions containing numbers in the base of the degree, 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 transformations are quite simple. In the vast majority of cases, they are based on the properties of the degree and are aimed mostly 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.

First, the exponents, in whose exponents the sum of some variable (or expression with variables) and a number, is found, 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 takes only positive values ​​on the ODZ variable x for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on subsequent transformations of expressions with powers ):

Now fractions with powers are cancelled, which gives .

Finally, the ratio of powers with the same exponents is replaced by powers of ratios, which leads to the equation , which is equivalent to . The transformations made allow us to introduce a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation

I. V. Boikov, L. D. Romanova Collection of tasks for preparing for the exam. Part 1. Penza 2003.

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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, a minus by a minus in this case gives a plus, and the sum does not change from a 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.

A literal expression (or an expression with variables) is a mathematical expression that consists of numbers, letters, and signs of mathematical operations. For example, the following expression is literal:

a+b+4

Using literal expressions, you can write down laws, formulas, equations, and functions. The ability to manipulate literal expressions is the key to a good knowledge of algebra and higher mathematics.

Any serious problem in mathematics comes down to solving equations. And to be able to solve equations, you need to be able to work with literal expressions.

To work with literal expressions, you need to study basic arithmetic well: addition, subtraction, multiplication, division, basic laws of mathematics, fractions, actions with fractions, proportions. And not just to study, but to understand thoroughly.

Lesson content

Variables

Letters that are contained in literal expressions are called variables. For example, in the expression a+b+ 4 variables are letters a And b. If instead of these variables we substitute any numbers, then the literal expression a+b+ 4 will turn into a numeric expression, the value of which can be found.

Numbers that are substituted for variables are called variable values. For example, let's change the values ​​of the variables a And b. Use the equals sign to change values

a = 2, b = 3

We have changed the values ​​of the variables a And b. variable a assigned a value 2 , variable b assigned a value 3 . As a result, the literal expression a+b+4 converts to a normal numeric expression 2+3+4 whose value can be found:

When variables are multiplied, they are written together. For example, the entry ab means the same as the entry a x b. If we substitute instead of variables a And b numbers 2 And 3 , then we get 6

Together, you can also write the multiplication of a number by an expression in brackets. For example, instead of a×(b + c) can be written a(b + c). Applying the distributive law of multiplication, we obtain a(b + c)=ab+ac.

Odds

In literal expressions, you can often find a notation in which a number and a variable are written together, for example 3a. In fact, this is a shorthand for multiplying the number 3 by a variable. a and this entry looks like 3×a .

In other words, the expression 3a is the product of the number 3 and the variable a. Number 3 in this work is called coefficient. This coefficient shows how many times the variable will be increased a. This expression can be read as " a three times or three times A", or "increment the value of the variable a three times", but most often read as "three a«

For example, if the variable a is equal to 5 , then the value of the expression 3a will be equal to 15.

3 x 5 = 15

talking plain language, the coefficient is the number that comes before the letter (before the variable).

There can be several letters, for example 5abc. Here the coefficient is the number 5 . This coefficient shows that the product of variables abc increases five times. This expression can be read as " abc five times" or "increase the value of the expression abc five times" or "five abc«.

If instead of variables abc substitute the numbers 2, 3 and 4, then the value of the expression 5abc will be equal to 120

5 x 2 x 3 x 4 = 120

You can mentally imagine how the numbers 2, 3 and 4 were first multiplied, and the resulting value increased five times:

The sign of the coefficient refers only to the coefficient, and does not apply to variables.

Consider the expression −6b. Minus in front of the coefficient 6 , applies only to the coefficient 6 , and does not apply to the variable b. Understanding this fact will allow you not to make mistakes in the future with signs.

Find the value of the expression −6b at b = 3.

−6b −6×b. For clarity, we write the expression −6b in expanded form and substitute the value of the variable b

−6b = −6 × b = −6 × 3 = −18

Example 2 Find the value of an expression −6b at b = −5

Let's write the expression −6b in expanded form

−6b = −6 × b = −6 × (−5) = 30

Example 3 Find the value of an expression −5a+b at a = 3 And b = 2

−5a+b is the short form for −5 × a + b, therefore, for clarity, we write the expression −5×a+b in expanded form and substitute the values ​​of the variables a And b

−5a + b = −5 × a + b = −5 × 3 + 2 = −15 + 2 = −13

Sometimes letters are written without a coefficient, for example a or ab. In this case, the coefficient is one:

but the unit is traditionally not written down, so they just write a or ab

If there is a minus before the letter, then the coefficient is a number −1 . For example, the expression -a actually looks like −1a. This is the product of minus one and the variable a. It came out like this:

−1 × a = −1a

Here lies a small trick. In the expression -a minus before variable a actually refers to the "invisible unit" and not the variable a. Therefore, when solving problems, you should be careful.

For example, given the expression -a and we are asked to find its value at a = 2, then at school we substituted a deuce instead of a variable a and get an answer −2 , not really focusing on how it turned out. In fact, there was a multiplication of minus one by a positive number 2

-a = -1 × a

−1 × a = −1 × 2 = −2

If an expression is given -a and it is required to find its value at a = −2, then we substitute −2 instead of a variable a

-a = -1 × a

−1 × a = −1 × (−2) = 2

In order to avoid mistakes, at first invisible units can be written explicitly.

Example 4 Find the value of an expression abc at a=2 , b=3 And c=4

Expression abc 1×a×b×c. For clarity, we write the expression abc a , b And c

1 x a x b x c = 1 x 2 x 3 x 4 = 24

Example 5 Find the value of an expression abc at a=−2 , b=−3 And c=−4

Let's write the expression abc in expanded form and substitute the values ​​of the variables a , b And c

1 × a × b × c = 1 × (−2) × (−3) × (−4) = −24

Example 6 Find the value of an expression − abc at a=3 , b=5 and c=7

Expression − abc is the short form for −1×a×b×c. For clarity, we write the expression − abc in expanded form and substitute the values ​​of the variables a , b And c

−abc = −1 × a × b × c = −1 × 3 × 5 × 7 = −105

Example 7 Find the value of an expression − abc at a=−2 , b=−4 and c=−3

Let's write the expression − abc expanded:

−abc = −1 × a × b × c

Substitute the value of the variables a , b And c

−abc = −1 × a × b × c = −1 × (−2) × (−4) × (−3) = 24

How to determine the coefficient

Sometimes it is required to solve a problem in which it is required to determine the coefficient of an expression. In principle, this task is very simple. It is enough to be able to correctly multiply numbers.

To determine the coefficient in an expression, you need to separately multiply the numbers included in this expression, and separately multiply the letters. The resulting numerical factor will be the coefficient.

Example 1 7m×5a×(−3)×n

The expression consists of several factors. This can be clearly seen if the expression is written in expanded form. That is, works 7m And 5a write in the form 7×m And 5×a

7 × m × 5 × a × (−3) × n

We apply the associative law of multiplication, which allows us to multiply factors in any order. Namely, separately multiply the numbers and separately multiply the letters (variables):

−3 × 7 × 5 × m × a × n = −105man

The coefficient is −105 . After completion, the letter part is preferably arranged in alphabetical order:

−105 am

Example 2 Determine the coefficient in the expression: −a×(−3)×2

−a × (−3) × 2 = −3 × 2 × (−a) = −6 × (−a) = 6a

The coefficient is 6.

Example 3 Determine the coefficient in the expression:

Let's multiply numbers and letters separately:

The coefficient is −1. Please note that the unit is not recorded, since the coefficient 1 is usually not recorded.

These seemingly simple tasks can play a very cruel joke with us. It often turns out that the sign of the coefficient is set incorrectly: either a minus is omitted or, on the contrary, it is set in vain. To avoid these annoying mistakes, it must be studied at a good level.

Terms in literal expressions

When you add several numbers, you get the sum of those numbers. Numbers that add up are called terms. There can be several terms, for example:

1 + 2 + 3 + 4 + 5

When an expression consists of terms, it is much easier to calculate it, since it is easier to add than to subtract. But the expression can contain not only addition, but also subtraction, for example:

1 + 2 − 3 + 4 − 5

In this expression, the numbers 3 and 5 are subtracted, not added. But nothing prevents us from replacing subtraction with addition. Then we again get an expression consisting of terms:

1 + 2 + (−3) + 4 + (−5)

It doesn't matter that the numbers -3 and -5 are now with a minus sign. The main thing is that all the numbers in this expression are connected by the addition sign, that is, the expression is a sum.

Both expressions 1 + 2 − 3 + 4 − 5 And 1 + 2 + (−3) + 4 + (−5) are equal to the same value - minus one

1 + 2 − 3 + 4 − 5 = −1

1 + 2 + (−3) + 4 + (−5) = −1

Thus, the value of the expression will not suffer from the fact that we replace subtraction with addition somewhere.

You can also replace subtraction with addition in literal expressions. For example, consider the following expression:

7a + 6b - 3c + 2d - 4s

7a + 6b + (−3c) + 2d + (−4s)

For any values ​​of variables a, b, c, d And s expressions 7a + 6b - 3c + 2d - 4s And 7a + 6b + (−3c) + 2d + (−4s) will be equal to the same value.

You must be prepared for the fact that a teacher at school or a teacher at an institute can call terms even those numbers (or variables) that are not them.

For example, if the difference is written on the board a-b, then the teacher will not say that a is the minuend, and b- deductible. He will call both variables one common word — terms. And all because the expression of the form a-b mathematician sees how the sum a + (−b). In this case, the expression becomes a sum, and the variables a And (−b) become components.

Similar terms

Similar terms are terms that have the same letter part. For example, consider the expression 7a + 6b + 2a. Terms 7a And 2a have the same letter part - variable a. So the terms 7a And 2a are similar.

Usually, like terms are added to simplify an expression or solve an equation. This operation is called reduction of like terms.

To bring like terms, you need to add the coefficients of these terms, and multiply the result by the common letter part.

For example, we give similar terms in the expression 3a + 4a + 5a. In this case, all terms are similar. We add their coefficients and multiply the result by the common letter part - by the variable a

3a + 4a + 5a = (3 + 4 + 5)×a = 12a

Such terms are usually given in the mind and the result is written immediately:

3a + 4a + 5a = 12a

Also, you can argue like this:

There were 3 variables a , 4 more variables a and 5 more variables a were added to them. As a result, we got 12 variables a

Let's consider several examples of reducing similar terms. Given that this topic very important, at first we will write down in detail every little thing. Despite the fact that everything is very simple here, most people make a lot of mistakes. Mostly due to inattention, not ignorance.

Example 1 3a + 2a + 6a + 8 a

We add the coefficients in this expression and multiply the result by the common letter part:

3a + 2a + 6a + 8a = (3 + 2 + 6 + 8) × a = 19a

design (3 + 2 + 6 + 8)×a you can not write down, so we will immediately write down the answer

3a + 2a + 6a + 8a = 19a

Example 2 Bring like terms in the expression 2a+a

Second term a written without a coefficient, but in fact it is preceded by a coefficient 1 , which we do not see due to the fact that it is not recorded. So the expression looks like this:

2a + 1a

Now we present similar terms. That is, we add the coefficients and multiply the result by the common letter part:

2a + 1a = (2 + 1) × a = 3a

Let's write the solution in short:

2a + a = 3a

2a+a, you can argue in another way:

Example 3 Bring like terms in the expression 2a - a

Let's replace subtraction with addition:

2a + (−a)

Second term (−a) written without a coefficient, but in fact it looks like (−1a). Coefficient −1 again invisible due to the fact that it is not recorded. So the expression looks like this:

2a + (−1a)

Now we present similar terms. We add the coefficients and multiply the result by the common letter part:

2a + (−1a) = (2 + (−1)) × a = 1a = a

Usually written shorter:

2a − a = a

Bringing like terms in the expression 2a−a You can also argue in another way:

There were 2 variables a , subtracted one variable a , as a result there was only one variable a

Example 4 Bring like terms in the expression 6a - 3a + 4a - 8a

6a − 3a + 4a − 8a = 6a + (−3a) + 4a + (−8a)

Now we present similar terms. We add the coefficients and multiply the result by the common letter part

(6 + (−3) + 4 + (−8)) × a = −1a = −a

Let's write the solution in short:

6a - 3a + 4a - 8a = -a

There are expressions that contain several various groups similar terms. For example, 3a + 3b + 7a + 2b. For such expressions, the same rules apply as for the rest, namely, adding the coefficients and multiplying the result by the common letter part. But in order to avoid mistakes, it is convenient to underline different groups of terms with different lines.

For example, in the expression 3a + 3b + 7a + 2b those terms that contain a variable a, can be underlined with one line, and those terms that contain a variable b, can be underlined with two lines:

Now we can bring like terms. That is, add the coefficients and multiply the result by the common letter part. This must be done for both groups of terms: for terms containing a variable a and for terms containing the variable b.

3a + 3b + 7a + 2b = (3+7)×a + (3 + 2)×b = 10a + 5b

Again, we repeat, the expression is simple, and similar terms can be given in the mind:

3a + 3b + 7a + 2b = 10a + 5b

Example 5 Bring like terms in the expression 5a - 6a - 7b + b

We replace subtraction with addition where possible:

5a − 6a −7b + b = 5a + (−6a) + (−7b) + b

Underline like terms with different lines. Terms containing variables a underline with one line, and the terms content are variables b, underlined with two lines:

Now we can bring like terms. That is, add the coefficients and multiply the result by the common letter part:

5a + (−6a) + (−7b) + b = (5 + (−6))×a + ((−7) + 1)×b = −a + (−6b)

If the expression contains ordinary numbers without alphabetic factors, then they are added separately.

Example 6 Bring like terms in the expression 4a + 3a − 5 + 2b + 7

Let's replace subtraction with addition where possible:

4a + 3a − 5 + 2b + 7 = 4a + 3a + (−5) + 2b + 7

Let us present similar terms. Numbers −5 And 7 do not have literal factors, but they are similar terms - you just need to add them up. And the term 2b will remain unchanged, since it is the only one in this expression that has a letter factor b, and there is nothing to add it with:

4a + 3a + (−5) + 2b + 7 = (4 + 3)×a + 2b + (−5) + 7 = 7a + 2b + 2

Let's write the solution in short:

4a + 3a − 5 + 2b + 7 = 7a + 2b + 2

Terms can be ordered so that those terms that have the same letter part are located in the same part of the expression.

Example 7 Bring like terms in the expression 5t+2x+3x+5t+x

Since the expression is the sum of several terms, this allows us to evaluate it in any order. Therefore, the terms containing the variable t, can be written at the beginning of the expression, and the terms containing the variable x at the end of the expression:

5t+5t+2x+3x+x

Now we can add like terms:

5t + 5t + 2x + 3x + x = (5+5)×t + (2+3+1)×x = 10t + 6x

Let's write the solution in short:

5t + 2x + 3x + 5t + x = 10t + 6x

The sum of opposite numbers is zero. This rule also works for literal expressions. If the expression contains the same terms, but with opposite signs, then you can get rid of them at the stage of reducing similar terms. In other words, just drop them from the expression because their sum is zero.

Example 8 Bring like terms in the expression 3t − 4t − 3t + 2t

Let's replace subtraction with addition where possible:

3t − 4t − 3t + 2t = 3t + (−4t) + (−3t) + 2t

Terms 3t And (−3t) are opposite. The sum of opposite terms is equal to zero. If we remove this zero from the expression, then the value of the expression will not change, so we will remove it. And we will remove it by the usual deletion of the terms 3t And (−3t)

As a result, we will have the expression (−4t) + 2t. In this expression, you can add like terms and get the final answer:

(−4t) + 2t = ((−4) + 2)×t = −2t

Let's write the solution in short:

Expression simplification

"simplify the expression" and the following is the expression to be simplified. Simplify Expression means to make it simpler and shorter.

In fact, we have already dealt with the simplification of expressions when reducing fractions. After the reduction, the fraction became shorter and easier to read.

Consider next example. Simplify the expression.

This task can be literally understood as follows: "Do whatever you can do with this expression, but make it simpler" .

In this case, you can reduce the fraction, namely, divide the numerator and denominator of the fraction by 2:

What else can be done? You can calculate the resulting fraction. Then we get the decimal 0.5

As a result, the fraction was simplified to 0.5.

The first question to ask yourself when solving such problems should be “what can be done?” . Because there are things you can do and there are things you can't do.

Another important point The thing to keep in mind is that the value of an expression must not change after the expression is simplified. Let's return to the expression. This expression is a division that can be performed. Having performed this division, we get the value of this expression, which is equal to 0.5

But we simplified the expression and got a new simplified expression . The value of the new simplified expression is still 0.5

But we also tried to simplify the expression by calculating it. As a result, the final answer was 0.5.

Thus, no matter how we simplify the expression, the value of the resulting expressions is still 0.5. This means that the simplification was carried out correctly at each stage. This is what we need to strive for when simplifying expressions - the meaning of the expression should not suffer from our actions.

It is often necessary to simplify literal expressions. For them, the same simplification rules apply as for numerical expressions. You can perform any valid action, as long as the value of the expression does not change.

Let's look at a few examples.

Example 1 Simplify Expression 5.21s × t × 2.5

To simplify this expression, you can multiply the numbers separately and multiply the letters separately. This task is very similar to the one we considered when we learned to determine the coefficient:

5.21s × t × 2.5 = 5.21 × 2.5 × s × t = 13.025 × st = 13.025st

So the expression 5.21s × t × 2.5 simplified to 13.025st.

Example 2 Simplify Expression −0.4×(−6.3b)×2

Second work (−6.3b) can be translated into a form understandable to us, namely, written in the form ( −6.3)×b , then separately multiply the numbers and separately multiply the letters:

− 0,4 × (−6.3b) × 2 = − 0,4 × (−6.3) × b × 2 = 5.04b

So the expression −0.4×(−6.3b)×2 simplified to 5.04b

Example 3 Simplify Expression

Let's write this expression in more detail in order to clearly see where the numbers are and where the letters are:

Now we multiply the numbers separately and multiply the letters separately:

So the expression simplified to −abc. This solution can be written shorter:

When simplifying expressions, fractions can be reduced in the process of solving, and not at the very end, as we did with ordinary fractions. For example, if in the course of solving we come across an expression of the form , then it is not at all necessary to calculate the numerator and denominator and do something like this:

A fraction can be reduced by choosing a factor in the numerator and in the denominator and reducing these factors by their largest common divisor. In other words, use , in which we do not describe in detail what the numerator and denominator were divided into.

For example, in the numerator, the factor 12 and in the denominator, the factor 4 can be reduced by 4. We keep the four in mind, and dividing 12 and 4 by this four, we write the answers next to these numbers, having previously crossed them out

Now you can multiply the resulting small factors. In this case, there are not many of them and you can multiply them in your mind:

Over time, you may find that when solving a particular problem, the expressions begin to “get fat”, so it is advisable to get used to fast calculations. What can be calculated in the mind must be calculated in the mind. What can be cut quickly should be cut quickly.

Example 4 Simplify Expression

So the expression simplified to

Example 5 Simplify Expression

We multiply numbers separately and letters separately:

So the expression simplified to mn.

Example 6 Simplify Expression

Let's write this expression in more detail in order to clearly see where the numbers are and where the letters are:

Now we multiply the numbers separately and the letters separately. For convenience of calculations, the decimal fraction −6.4 and the mixed number can be converted to ordinary fractions:

So the expression simplified to

The solution for this example can be written much shorter. It will look like this:

Example 7 Simplify Expression

We multiply numbers separately and letters separately. For ease of calculation, the mixed number and decimals 0.1 and 0.6 can be converted to ordinary fractions:

So the expression simplified to abcd. If you skip the details, then this solution can be written much shorter:

Notice how the fraction has been reduced. New multipliers, which are obtained by reducing the previous multipliers, can also be reduced.

Now let's talk about what not to do. When simplifying expressions, it is strictly forbidden to multiply numbers and letters if the expression is a sum and not a product.

For example, if you want to simplify the expression 5a + 4b, then it cannot be written as follows:

This is equivalent to the fact that if we were asked to add two numbers, and we would multiply them instead of adding them.

When substituting any values ​​of variables a And b expression 5a+4b turns into a simple numeric expression. Let's assume the variables a And b have the following meanings:

a = 2 , b = 3

Then the value of the expression will be 22

5a + 4b = 5 × 2 + 4 × 3 = 10 + 12 = 22

First, the multiplication is performed, and then the results are added. And if we tried to simplify this expression by multiplying numbers and letters, we would get the following:

5a + 4b = 5 × 4 × a × b = 20ab

20ab = 20 x 2 x 3 = 120

It turns out a completely different meaning of the expression. In the first case it turned out 22 , in the second case 120 . This means that the simplification of the expression 5a + 4b was performed incorrectly.

After simplifying the expression, its value should not change with the same values ​​of the variables. If, when substituting any variable values ​​into the original expression, one value is obtained, then after simplifying the expression, the same value should be obtained as before simplification.

With expression 5a + 4b actually nothing can be done. It doesn't get easier.

If the expression contains similar terms, then they can be added if our goal is to simplify the expression.

Example 8 Simplify Expression 0.3a−0.4a+a

0.3a − 0.4a + a = 0.3a + (−0.4a) + a = (0.3 + (−0.4) + 1)×a = 0.9a

or shorter: 0.3a - 0.4a + a = 0.9a

So the expression 0.3a−0.4a+a simplified to 0.9a

Example 9 Simplify Expression −7.5a − 2.5b + 4a

To simplify this expression, you can add like terms:

−7.5a − 2.5b + 4a = −7.5a + (−2.5b) + 4a = ((−7.5) + 4)×a + (−2.5b) = −3.5a + (−2.5b)

or shorter −7.5a − 2.5b + 4a = −3.5a + (−2.5b)

term (−2.5b) remained unchanged, since there was nothing to fold it with.

Example 10 Simplify Expression

To simplify this expression, you can add like terms:

The coefficient was for the convenience of calculation.

So the expression simplified to

Example 11. Simplify Expression

To simplify this expression, you can add like terms:

So the expression simplified to .

IN this example it would make more sense to add the first and last coefficient first. In this case, we would get a short solution. It would look like this:

Example 12. Simplify Expression

To simplify this expression, you can add like terms:

So the expression simplified to .

The term remained unchanged, since there was nothing to add it to.

This solution can be written much shorter. It will look like this:

The short solution omits the steps of replacing subtraction with addition and a detailed record of how the fractions were reduced to a common denominator.

Another difference is that in detailed solution the answer looks like , but in short as . Actually, it's the same expression. The difference is that in the first case, subtraction is replaced by addition, since at the beginning when we wrote the solution in detailed view, we have replaced subtraction with addition wherever possible, and this replacement has been preserved for the answer.

Identities. Identical equal expressions

After we have simplified any expression, it becomes simpler and shorter. To check whether an expression is simplified correctly, it is enough to substitute any values ​​of the variables first into the previous expression, which was to be simplified, and then into the new one, which was simplified. If the value in both expressions is the same, then the expression is simplified correctly.

Consider the simplest example. Let it be required to simplify the expression 2a × 7b. To simplify this expression, you can separately multiply the numbers and letters:

2a × 7b = 2 × 7 × a × b = 14ab

Let's check if we simplified the expression correctly. To do this, substitute any values ​​of the variables a And b first to the first expression, which needed to be simplified, and then to the second, which was simplified.

Let the values ​​of the variables a , b will be as follows:

a = 4 , b = 5

Substitute them in the first expression 2a × 7b

Now let's substitute the same values ​​of the variables into the expression that resulted from the simplification 2a×7b, namely in the expression 14ab

14ab = 14 x 4 x 5 = 280

We see that at a=4 And b=5 the value of the first expression 2a×7b and the value of the second expression 14ab equal

2a × 7b = 2 × 4 × 7 × 5 = 280

14ab = 14 x 4 x 5 = 280

The same will happen for any other values. For example, let a=1 And b=2

2a × 7b = 2 × 1 × 7 × 2 = 28

14ab = 14 x 1 x 2 = 28

Thus, for any values ​​of the variables, the expressions 2a×7b And 14ab are equal to the same value. Such expressions are called identically equal.

We conclude that between the expressions 2a×7b And 14ab you can put an equal sign, since they are equal to the same value.

2a × 7b = 14ab

An equality is any expression that is joined by an equal sign (=).

And the equality of the form 2a×7b = 14ab called identity.

An identity is an equality that is true for any values ​​of the variables.

Other examples of identities:

a + b = b + a

a(b+c) = ab + ac

a(bc) = (ab)c

Yes, the laws of mathematics that we studied are identities.

True numerical equalities are also identities. For example:

2 + 2 = 4

3 + 3 = 5 + 1

10 = 7 + 2 + 1

Solving a complex problem to make it easier for yourself to calculate, compound expression replaced by a simpler expression, identically equal to the previous one. Such a replacement is called identical transformation of the expression or simply expression conversion.

For example, we simplified the expression 2a × 7b, and get a simpler expression 14ab. This simplification can be called the identity transformation.

You can often find a task that says "prove that equality is identity" and then the equality to be proved is given. Usually this equality consists of two parts: the left and right parts of the equality. Our task is to perform identical transformations with one of the parts of the equality and get the other part. Or perform identical transformations with both parts of the equality and make sure that both parts of the equality contain the same expressions.

For example, let us prove that the equality 0.5a × 5b = 2.5ab is an identity.

Simplify the left side of this equality. To do this, multiply the numbers and letters separately:

0.5 × 5 × a × b = 2.5ab

2.5ab = 2.5ab

As a result of a small identical transformation, left side equality became equal to the right side of the equality. So we have proved that the equality 0.5a × 5b = 2.5ab is an identity.

From identical transformations, we learned to add, subtract, multiply and divide numbers, reduce fractions, bring like terms, and also simplify some expressions.

But these are far from all identical transformations that exist in mathematics. There are many more identical transformations. We will see this again and again in the future.

Tasks for independent solution:

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