What is an integer expression rule. Lesson "Algebraic fractions, rational and fractional expressions

An integer expression is a mathematical expression made up of numbers and literal variables using the operations of addition, subtraction, and multiplication. Integers also include expressions that include division by some number other than zero.

Integer Expression Examples

Below are some examples of integer expressions:

1. 12*a^3 + 5*(2*a -1);

2.7*b

3. 4*y- ((5*y+3)/5) -1;

Fractional Expressions

If the expression contains a division by a variable or by another expression containing a variable, then such an expression is not an integer. Such an expression is called a fractional expression. Let us give a complete definition of a fractional expression.

A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed with numbers and literal variables, as well as division by a number not equal to zero, also contains division into expressions with literal variables.

Examples of fractional expressions:

1. (12*a^3 +4)/a

2.7/(x+3)

3. 4*x- ((5*y+3)/(5-y)) +1;

Fractional and integer expressions make up two large sets of mathematical expressions. If these sets are combined, then we get a new set, which is called rational expressions. That is, rational expressions are all integer and fractional expressions.

We know that integer expressions make sense for any values of the variables that are included in it. This follows from the fact that in order to find the value of an integer expression, it is necessary to perform actions that are always possible: addition, subtraction, multiplication, division by a number other than zero.

Fractional expressions, unlike integer ones, may not make sense. Since there is a division operation by a variable or an expression containing variables, and this expression can turn into zero, but you cannot divide by zero. Variable values for which the fractional expression will make sense are called valid variable values.

rational fraction

One of the special cases of rational expressions will be a fraction, the numerator and denominator of which are polynomials. For such a fraction in mathematics, there is also a name - a rational fraction.

A rational fraction will make sense if its denominator is not equal to zero. That is, all values of variables for which the denominator of the fraction is different from zero will be valid.

Integer Expression Examples

Below are some examples of integer expressions:

1. 12*a^3 + 5*(2*a -1);

3. 4*y- ((5*y+3)/5) -1;

Fractional Expressions

Examples of fractional expressions:

1. (12*a^3 +4)/a

3. 4*x- ((5*y+3)/(5-y)) +1;

rational fraction

A rational fraction will make sense if its denominator is not equal to zero. That is, all values of variables for which the denominator of the fraction is different from zero will be valid.

“Polynomial Lesson” - And check: 2. Perform the multiplication of polynomials: 4. Perform the division of the polynomial A (x) by B (x). 3. Factorize the polynomial. 1. Perform addition and subtraction of polynomials: P(x)=-2x3 + x2 -x-12 and Q(x)= x3 -3x2 -4x+1. Actions with polynomials. Lesson 15

"Converting an integer expression to a polynomial" - Develop students' computational skills. Introduce the concept of a whole expression. Converting integer expressions. Polynomials and, in particular, monomials are integer expressions. Exercise students in bringing like terms. Examples of integer expressions are: 10y?+(3x+y)(x?-10y?), 2b(b?-10c?)-(b?+2c?), 3a?-(a(a+2c) )/5+2.5ac.

"Polynomial multiplication" - -x6+3x7-2x4+5x2 3 -1 0 -2 0 5 0 0 7 -8 3 5 -6 7x4-8x3+3x2+5x-6. Presentation. The positional number of a polynomial. Multiplication of polynomials using a positional number. Ryabov Pavel Yurievich. Head: Kaleturina A.S.

"Standard form polynomial" - The standard form of a polynomial. Examples. 3x4 + 2x3 - x2 + 5. Addition of polynomials. Preparation for s / r No. 6. Vocabulary. Chapter 2, §1b. For polynomials with one letter, the leading term is uniquely defined. Check yourself. 6x4 - x3y + x2y2 + 2y4.

"Polynomials" - A monomial is considered a polynomial consisting of one member. Taking the common factor out of brackets. Algebra. Polynomials. Multiply the polynomial a+b by the polynomial c+d. Product of a monomial and a polynomial Multiplication of a monomial by a polynomial. Similar terms are members 2 and -7, which do not have a letter part. The terms of the polynomial 4xz-5xy+3x-1 are 4xz, -5xy, 3x and -1.

"Lesson Factoring" - Application of FSU. Abbreviated multiplication formulas. Lesson topic: Answers: var 1: b, d, b, d, c; var 2: a, d, c, b, a; var 3: c, c, c, a, b; Option 4: d, d, c, b, d. So how? Taking the common factor out of brackets. 3. Complete the factorization: Group work: Put the common factor out of brackets. 1. Finish the factorization: a).

Thanks to the algebra course, it is known that all expressions require transformation for a more convenient solution. Defining integer expressions encourages identical transformations to begin with. We will transform the expression into a polynomial. In conclusion, let's look at a few examples.

Definition and examples of integer expressions

Definition 1

Integer expressions are numbers, variables, or expressions with addition or subtraction, which are written as a power with a natural exponent, which also have brackets or division other than zero.

Based on the definition, we have that examples of integer expressions: 7 , 0 , − 12 , 7 11 , 2 , 73 , - 3 5 6 and so on, and variables of the form a , b , p , q , x , z are considered integers expressions. After their transformation of sums, differences, products, the expressions will take the form

x + 1 , 5 y 3 2 3 7 − 2 y − 3 , 3 − x y z 4 , - 6 7 , 5 (2 x + 3 y 2) 2 − - ( 1 − x) (1 + x) (1 + x 2)

If the expression contains a division by a number other than zero of the form x: 5 + 8: 2: 4 or (x + y) : 6 , then the division can be indicated with a slash, as x + 3 5 - 3 , 2 x + 2 . When considering expressions of the form x: 5 + 5: x or 4 + a 2 + 2 a - 6 a + b + 2 c, it is clear that such expressions cannot be integers, since in the first there is a division by the variable x, and in the second to an expression with a variable.

Polynomial and monomial are integer expressions that we meet at school when working with rational numbers. In other words, integer expressions do not include irrational fractions. Another name is whole irrational expressions.

What transformations of integer expressions are possible?

Integer expressions are considered when solving as basic identical transformations, opening brackets, grouping, reduction of similar ones.

Example 1

Open the brackets and bring like terms into 2 · (a 3 + 3 · a · b − 2 · a) − 2 · a 3 − (5 · a · b − 6 · a + b) .

Decision

First you need to apply the rule for opening brackets. We get an expression of the form 2 (a 3 + 3 a b − 2 a) − 2 a 3 − (5 a b − 6 a + b) = = 2 a 3 + 2 3 a b + 2 (− 2 a) − 2 a 3 − 5 a b + 6 a − b = = 2 a 3 + 6 a b − 4 a − 2 a 3 − 5 a b + 6 a − b

Then we can add like terms:

2 a 3 + 6 a b − 4 a − 2 a 3 − 5 a b + 6 a − b = = (2 a 3 − 2 a 3) + (6 a b − 5 a b) + (− 4 a + 6 a) − b = = 0 + a b + 2 a − b = a b + 2 a − b .

After reducing them, we obtain a polynomial of the form a · b + 2 · a − b .

Answer: 2 (a 3 + 3 a b − 2 a) − 2 a 3 − (5 a b − 6 a + b) = a b + 2 a − b.

Example 2

Make transformations (x - 1) : 2 3 + 2 · (x 2 + 1) : 3: 7 .

Decision

The existing division can be replaced by multiplication, but by the reciprocal of the number. Then it is necessary to perform transformations, after which the expression will take the form (x - 1) · 3 2 + 2 · (x 2 + 1) · 1 3 · 1 7 . Now we should deal with the reduction of like terms. We get that

(x - 1) 3 2 + 2 (x 2 + 1) 1 3 1 7 = 3 2 (x - 1) + 2 21 x 2 + 1 = = 3 2 x - 3 2 + 2 21 x 2 + 2 21 = 2 21 x 2 + 3 2 x - 59 42 = 2 21 x 2 + 1 1 2 x - 1 17 42

Answer: (x - 1) : 2 3 + 2 (x 2 + 1) : 3: 7 = 2 21 x 2 + 1 1 2 x - 1 17 42 .

Example 3

Express the expression 6 x 2 y + 18 x y − 6 y − (x 2 + 3 x − 1) (x 3 + 4 x) as a product.

Decision

Having examined the expression, it is clear that the first three terms have a common factor of the form 6 · y , which should be taken out of brackets during the transformation. Then we get that 6 x 2 y + 18 x y − 6 y − (x 2 + 3 x − 1) (x 3 + 4 x) = = 6 y (x 2 + 3 x − 1) - (x 2 + 3 x - 1) (x 3 + 4 x)

It can be seen that we obtained the difference of two expressions of the form 6 y (x 2 + 3 x - 1) and (x 2 + 3 x - 1) (x 3 + 4 x) with a common factor x 2 + 3 x − 1 , which must be taken out of brackets. We get that

6 y (x 2 + 3 x − 1) − (x 2 + 3 x − 1) (x 3 + 4 x) = = (x 2 + 3 x − 1) (6 y − (x 3 + 4 x))

Having opened the brackets, we have an expression of the form (x 2 + 3 x - 1) (6 y - x 3 - 4 x) , which had to be found by condition.

Answer:6 x 2 y + 18 x y − 6 y − (x 2 + 3 x − 1) (x 3 + 4 x) = = (x 2 + 3 x − 1) ( 6 y − x 3 − 4 x)

Identical transformations require strict implementation of the order of operations.

Example 4

Convert expression (3 2 − 6 2: 9) 3 (x 2) 4 + 4 x: 8.

Decision

You first perform the actions in parentheses. Then we have that 3 2 - 6 2: 9 = 3 2 - 3 6: 9 = 6 - 4 = 2. After transformations, the expression becomes 2 3 · (x 2) 4 + 4 · x: 8 . It is known that 2 3 = 8 and (x 2) 4 = x 2 4 = x 8, then you can come to an expression like 8 x 8 + 4 x: 8 . The second term requires the replacement of division by multiplication from 4x:8. Grouping the factors, we get that

8 x 8 + 4 x: 8 = 8 x 8 + 4 x 1 8 = 8 x 8 + 4 1 8 x = 8 x 8 + 1 2 x

Answer:(3 2 − 6 2: 9) 3 (x 2) 4 + 4 x: 8 = 8 x 8 + 1 2 x .

Polynomial conversion

Most of the conversions of integer expressions are polynomial representations. Any expression can be represented as a polynomial. Any expression can be considered as polynomials connected by arithmetic signs. Any operation on polynomials results in a polynomial.

In order for the expression to be represented as a polynomial, it is necessary to perform all actions with polynomials, according to the algorithm.

Example 5

Express as a polynomial 2 (2 x 3 - 1) + (2 x - 1) 2 (3 - x) + (4 x - x (15 x + 1)) .

Decision

In this expression, start the transformations with an expression of the form 4 x − x (15 x + 1) , and according to the rule, at the beginning by performing multiplication or division, after which addition or subtraction. Multiply - x by 15 x + 1, then we get 4 x - x (15 x + 1) = 4 x - 15 x 2 - x = (4 x - x) - 15 x 2 = 3 x - 15 x 2. The given expression will take the form 2 (2 x 3 - 1) + (2 x - 1) 2 (3 - x) + (3 x - 15 x 2) .

Next, you need to raise the polynomial to the 2nd power 2x-1, we get an expression of the form (2 x − 1) 2 = (2 x − 1) (2 x − 1) = 4 x 2 + 2 x (− 1) − 1 2 x − 1 (− 1 ) = = 4 x 2 − 4 x + 1

Now we can go to the view 2 (2 x 3 - 1) + (4 x 2 - 4 x + 1) (3 - x) + (3 x - 15 x 2).

Let's look at multiplication. It can be seen that 2 (2 x 3 - 1) = 4 x 3 - 2 and (4 x 2 - 4 x + 1) (3 - x) = 12 x 2 - 4 x 3 - 12 x + 4 x 2 + 3 - x = = 16 x 2 - 4 x 3 - 13 x + 3

then you can make a transition to an expression of the form (4 x 3 - 2) + (16 x 2 - 4 x 3 - 13 x + 3) + (3 x - 15 x 2).

We perform addition, after which we arrive at the expression:

(4 x 3 - 2) + (16 x 2 - 4 x 3 - 13 x + 3) + (3 x - 15 x 2) = = 4 x 3 - 2 + 16 x 2 − 4 x 3 − 13 x + 3 + 3 x − 15 x 2 = = (4 x 3 − 4 x 3) + (16 x 2 − 15 x 2) + (− 13 x + 3 x) + (− 2 + 3) = = 0 + x 2 − 10 x + 1 = x 2 − 10 x + 1 .

It follows that the original expression has the form x 2 − 10 x + 1.

Answer: 2 (2 x 3 - 1) + (2 x - 1) 2 (3 - x) + (4 x - x (15 x + 1)) = x 2 - 10 x + 1.

Multiplication and exponentiation of a polynomial indicates that it is necessary to use abbreviated multiplication formulas to speed up the conversion process. This contributes to the fact that the actions will be performed rationally and correctly.

Example 6

Convert 4 · (2 · m + n) 2 + (m − 2 · n) · (m + 2 · n) .

Decision

From the square formula, we get that (2 m + n) 2 = (2 m) 2 + 2 (2 m) n + n 2 = 4 m 2 + 4 m n + n 2, then the product (m − 2 n) (m + 2 n) equals the difference of the squares m and 2 n , thus equals m 2 − 4 n 2. We get that the original expression takes the form 4 (2 m + n) 2 + (m − 2 n) (m + 2 n) = 4 (4 m 2 + 4 m n + n 2) + (m 2 − 4 n 2) = = 16 m 2 + 16 m n + 4 n 2 + m 2 − 4 n 2 = 17 m 2 + 16 m n

Answer: 4 (2 m + n) 2 + (m − 2 n) (m + 2 n) = 17 m 2 + 16 m n.

In order for the transformation not to be too long, it is necessary to bring the given expression to the standard form.

Example 7

Simplify the expression (2 a (− 3) a 2 b) (2 a + 5 b 2) + a b (a 2 + 1 + a 2) (6 a + 15 b 2 ) + (5 a b (− 3) b 2)

Decision

Most often, polynomials and monomials are not given in a standard form, so you have to perform transformations. Should be converted to get an expression of the form − 6 a 3 b (2 a + 5 b 2) + a b (2 a 2 + 1) (6 a + 15 b 2) − 15 a b 3. In order to bring similar ones, it is necessary to first perform multiplication according to the rules for transforming a complex expression. We get an expression like

− 6 a 3 b (2 a + 5 b 2) + a b (2 a 2 + 1) (6 a + 15 b 2) − 15 a b 3 = = − 12 a 4 b − 30 a 3 b 3 + (2 a 3 b + a b) (6 a + 15 b 2) − 15 a b 3 = = − 12 a 4 b − 30 a 3 b 3 + 12 a 4 b + 30 a 3 b 3 + 6 a 2 b + 15 a b 3 − 15 a b 3 = = (− 12 a 4 b + 12 a 4 b) + (− 30 a 3 b 3 + 30 a 3 b 3) + 6 a 2 b + (15 a b 3 − 15 a b 3) = 6 a 2 b

Answer: (2 a (− 3) a 2 b) (2 a + 5 b 2) + a b (a 2 + 1 + a 2) (6 a + 15 b 2 ) + + (5 a b (− 3) b 2) = 6 a 2 b

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