The rule for opening brackets in multiplication. Bracket opening: rules and examples (Grade 7)

In this lesson, you will learn how to transform an expression that contains parentheses into an expression that does not contain parentheses. You will learn how to open brackets preceded by a plus sign and a minus sign. We will remember how to open brackets using the distributive law of multiplication. The considered examples will allow linking new and previously studied material into a single whole.

Topic: Equation Solving

Lesson: Parentheses expansion

How to open brackets preceded by a "+" sign. Use of the associative law of addition.

If you need to add the sum of two numbers to a number, then you can add the first term to this number, and then the second.

To the left of the equal sign is an expression with parentheses, and to the right is an expression without parentheses. This means that when passing from the left side of the equality to the right side, the brackets were opened.

Consider examples.

Example 1

Expanding the brackets, we changed the order of operations. Counting has become more convenient.

Example 2

Example 3

Note that in all three examples, we simply removed the parentheses. Let's formulate the rule:

Comment.

If the first term in brackets is unsigned, then it must be written with a plus sign.

You can follow the step by step example. First, add 445 to 889. This mental action can be performed, but it is not very easy. Let's open the brackets and see that the changed order of operations will greatly simplify the calculations.

If you follow the indicated order of actions, then you must first subtract 345 from 512, and then add 1345 to the result. By expanding the brackets, we will change the order of actions and greatly simplify the calculations.

Illustrative example and rule.

Consider an example: . You can find the value of the expression by adding 2 and 5, and then taking the resulting number with the opposite sign. We get -7.

On the other hand, the same result can be obtained by adding the opposite numbers.

Let's formulate the rule:

Example 1

Example 2

The rule does not change if there are not two, but three or more terms in brackets.

Example 3

Comment. Signs are reversed only in front of the terms.

In order to open the brackets, in this case, we need to recall the distributive property.

First, multiply the first bracket by 2 and the second by 3.

The first bracket is preceded by a “+” sign, which means that the signs must be left unchanged. The second is preceded by a “-” sign, therefore, all signs must be reversed

Bibliography

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012.
2. Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium, 2006.
3. Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - Enlightenment, 1989.
4. Rurukin A.N., Tchaikovsky I.V. Tasks for the course of mathematics grade 5-6 - ZSH MEPhI, 2011.
5. Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for students of the 6th grade of the MEPhI correspondence school. - ZSH MEPhI, 2011.
6. Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Interlocutor textbook for grades 5-6 high school. Library of the teacher of mathematics. - Enlightenment, 1989.

1. Online math tests ().
2. You can download the ones specified in clause 1.2. books().

Homework

1. Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M .: Mnemosyne, 2012. (see link 1.2)
2. Homework: No. 1254, No. 1255, No. 1256 (b, d)
3. Other assignments: No. 1258(c), No. 1248

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The main function of brackets is to change the order of actions when calculating values. For example, in the numerical expression \(5 3+7\) the multiplication will be calculated first, and then the addition: \(5 3+7 =15+7=22\). But in the expression \(5·(3+7)\), the addition in the bracket will be calculated first, and only then the multiplication: \(5·(3+7)=5·10=50\).

Example. Expand the bracket: \(-(4m+3)\).
Solution : \(-(4m+3)=-4m-3\).

Example. Expand the bracket and give like terms \(5-(3x+2)+(2+3x)\).
Solution : \(5-(3x+2)+(2+3x)=5-3x-2+2+3x=5\).

Example. Expand the brackets \(5(3-x)\).
Solution : We have \(3\) and \(-x\) in the bracket, and five in front of the bracket. This means that each member of the bracket is multiplied by \ (5 \) - I remind you that the multiplication sign between a number and a bracket in mathematics is not written to reduce the size of records.

Example. Expand the brackets \(-2(-3x+5)\).
Solution : As in the previous example, the bracketed \(-3x\) and \(5\) are multiplied by \(-2\).

Example. Simplify the expression: \(5(x+y)-2(x-y)\).
Solution : \(5(x+y)-2(x-y)=5x+5y-2x+2y=3x+7y\).

It remains to consider the last situation.

When multiplying parenthesis by parenthesis, each term of the first parenthesis is multiplied with every term of the second:

\((c+d)(a-b)=c (a-b)+d (a-b)=ca-cb+da-db\)

Example. Expand the brackets \((2-x)(3x-1)\).
Solution : We have a product of brackets and it can be opened immediately using the formula above. But in order not to get confused, let's do everything step by step.
Step 1. Remove the first bracket - each of its members is multiplied by the second bracket:

Step 2. Expand the products of the bracket by the factor as described above:
- the first one first...

Then the second.

Step 3. Now we multiply and bring like terms:

It is not necessary to paint all the transformations in detail, you can immediately multiply. But if you are just learning to open brackets - write in detail, there will be less chance of making a mistake.

Note to the entire section. In fact, you don't need to remember all four rules, you only need to remember one, this one: \(c(a-b)=ca-cb\) . Why? Because if we substitute one instead of c, we get the rule \((a-b)=a-b\) . And if we substitute minus one, we get the rule \(-(a-b)=-a+b\) . Well, if you substitute another bracket instead of c, you can get the last rule.

parenthesis within parenthesis

Sometimes in practice there are problems with brackets nested inside other brackets. Here is an example of such a task: to simplify the expression \(7x+2(5-(3x+y))\).

To be successful in these tasks, you need to:
- carefully understand the nesting of brackets - which one is in which;
- open the brackets sequentially, starting, for example, with the innermost one.

It is important when opening one of the brackets don't touch the rest of the expression, just rewriting it as is.
Let's take the task above as an example.

Example. Open the brackets and give like terms \(7x+2(5-(3x+y))\).
Solution:

Example. Expand the brackets and give like terms \(-(x+3(2x-1+(x-5)))\).
Solution :

\(-(x+3(2x-1\)\(+(x-5)\) \())\)		This is a triple nesting of parentheses. We start with the innermost one (highlighted in green). There is a plus in front of the parenthesis, so it is simply removed.
\(-(x+3(2x-1\)\(+x-5\) \())\)		Now you need to open the second bracket, intermediate. But before that, we will simplify the expression by ghosting similar terms in this second bracket.
\(=-(x\)\(+3(3x-6)\) \()=\)		Now we open the second bracket (highlighted in blue). There is a multiplier in front of the parenthesis - so each term in the parenthesis is multiplied by it.
\(=-(x\)\(+9x-18\) \()=\)
		And open the last parenthesis. Before the bracket minus - so all the signs are reversed.

Bracket opening is a basic skill in mathematics. Without this skill, it is impossible to have a grade above three in grades 8 and 9. Therefore, I recommend a good understanding of this topic.

A + (b + c) can be written without brackets: a + (b + c) \u003d a + b + c. This operation is called parenthesis expansion.

Example 1 Let's open the brackets in the expression a + (- b + c).

Solution. a + (-b + c) = a + ((-b) + c) = a + (-b) + c = a-b + c.

If there is a “+” sign before the brackets, then you can omit the brackets and this “+” sign, retaining the signs of the terms in the brackets. If the first term in brackets is written without a sign, then it must be written with a “+” sign.

Example 2 Let's find the value of the expression -2.87+ (2.87-7.639).

Solution. Opening the brackets, we get - 2.87 + (2.87 - 7.639) \u003d - - 2.87 + 2.87 - 7.639 \u003d 0 - 7.639 \u003d - 7.639.

To find the value of the expression - (- 9 + 5), you need to add numbers-9 and 5 and find the number opposite to the amount received: -(- 9 + 5)= -(- 4) = 4.

The same value can be obtained in a different way: first write down the numbers opposite to these terms (i.e. change their signs), and then add: 9 + (- 5) = 4. Thus, - (- 9 + 5) = 9 - 5 = 4.

To write the sum opposite to the sum of several terms, it is necessary to change the signs of these terms.

So - (a + b) \u003d - a - b.

Example 3 Find the value of the expression 16 - (10 -18 + 12).

Solution. 16-(10 -18 + 12) = 16 + (-(10 -18 + 12)) = = 16 + (-10 +18-12) = 16-10 +18-12 = 12.

To open the brackets preceded by the “-” sign, you need to replace this sign with “+”, changing the signs of all the terms in the brackets to the opposite ones, and then open the brackets.

Example 4 Let's find the value of the expression 9.36-(9.36 - 5.48).

Solution. 9.36 - (9.36 - 5.48) = 9.36 + (- 9.36 + 5.48) == 9.36 - 9.36 + 5.48 = 0 -f 5.48 = 5 .48.

Bracket opening and the use of commutative and associative properties additions make calculations easier.

Example 5 Find the value of the expression (-4-20)+(6+13)-(7-8)-5.

Solution. First, we open the brackets, and then we find separately the sum of all positive and separately the sum of all negative numbers, and finally add the results:

(- 4 - 20)+(6+ 13)-(7 - 8) - 5 = -4-20 + 6 + 13-7 + 8-5 = = (6 + 13 + 8)+(- 4 - 20 - 7 - 5)= 27-36=-9.

Example 6 Find the value of the expression

Solution. First, we represent each term as the sum of their integer and fractional parts, then open the brackets, then add the whole and separately fractional parts and finally sum up the results:

How do you open parentheses that are preceded by a "+" sign? How can you find the value of an expression that is the opposite of the sum of several numbers? How to open brackets preceded by a "-" sign?

1218. Expand the brackets:

a) 3.4+(2.6+ 8.3); c) m+(n-k);

b) 4.57+(2.6 - 4.57); d) c+(-a + b).

1219. Find the value of the expression:

1220. Expand the brackets:

a) 85+(7.8+ 98); d) -(80-16) + 84; g) a-(b-k-n);
b) (4.7 -17) + 7.5; e) -a + (m-2.6); h) - (a-b + c);
c) 64-(90 + 100); e) c+(-a-b); i) (m-n)-(p-k).

1221. Expand the brackets and find the value of the expression:

1222. Simplify the expression:

1223. Write amount two expressions and simplify it:

a) - 4 - m and m + 6.4; d) a + b and p - b
b) 1.1+a and -26-a; e) - m + n and -k - n;
c) a + 13 and -13 + b; e)m - n and n - m.

1224. Write the difference of two expressions and simplify it:

1226. Use the equation to solve the problem:

a) There are 42 books on one shelf, and 34 on the other. Several books were removed from the second shelf, and as many as were left on the second from the first. After that, 12 books remained on the first shelf. How many books were taken off the second shelf?

b) There are 42 students in the first class, 3 students less in the second than in the third. How many students are in the third grade if there are 125 students in these three grades?

1227. Find the value of the expression:

1228. Calculate orally:

1229. Find highest value expressions:

1230. Enter 4 consecutive integers if:

a) the smaller of them is equal to -12; c) the smaller of them is equal to n;
b) the greater of them is equal to -18; d) the larger of them is equal to k.

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Among the various expressions that are considered in algebra, sums of monomials occupy an important place. Here are examples of such expressions:
\(5a^4 - 2a^3 + 0.3a^2 - 4.6a + 8 \)
\(xy^3 - 5x^2y + 9x^3 - 7y^2 + 6x + 5y - 2 \)

The sum of monomials is called a polynomial. The terms in a polynomial are called members of the polynomial. Mononomials are also referred to as polynomials, considering a monomial as a polynomial consisting of one member.

For example, polynomial
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 \)
can be simplified.

We represent all the terms as monomials of the standard form:
\(8b^5 - 2b \cdot 7b^4 + 3b^2 - 8b + 0.25b \cdot (-12)b + 16 = \)
\(= 8b^5 - 14b^5 + 3b^2 -8b -3b^2 + 16 \)

We give similar terms in the resulting polynomial:
\(8b^5 -14b^5 +3b^2 -8b -3b^2 + 16 = -6b^5 -8b + 16 \)
The result is a polynomial, all members of which are monomials of the standard form, and among them there are no similar ones. Such polynomials are called polynomials of standard form.

Behind polynomial degree standard form take the largest of the powers of its members. So, the binomial \(12a^2b - 7b \) has the third degree, and the trinomial \(2b^2 -7b + 6 \) has the second.

Usually, the members of standard form polynomials containing one variable are arranged in descending order of its exponents. For example:
\(5x - 18x^3 + 1 + x^5 = x^5 - 18x^3 + 5x + 1 \)

The sum of several polynomials can be converted (simplified) into a standard form polynomial.

Sometimes the members of a polynomial need to be divided into groups, enclosing each group in parentheses. Since parentheses are the opposite of parentheses, it is easy to formulate parentheses opening rules:

If the + sign is placed before the brackets, then the terms enclosed in brackets are written with the same signs.

If a "-" sign is placed in front of the brackets, then the terms enclosed in brackets are written with opposite signs.

Transformation (simplification) of the product of a monomial and a polynomial

Using the distributive property of multiplication, one can transform (simplify) the product of a monomial and a polynomial into a polynomial. For example:
\(9a^2b(7a^2 - 5ab - 4b^2) = \)
\(= 9a^2b \cdot 7a^2 + 9a^2b \cdot (-5ab) + 9a^2b \cdot (-4b^2) = \)
\(= 63a^4b - 45a^3b^2 - 36a^2b^3 \)

The product of a monomial and a polynomial is identically equal to the sum of the products of this monomial and each of the terms of the polynomial.

This result is usually formulated as a rule.

To multiply a monomial by a polynomial, one must multiply this monomial by each of the terms of the polynomial.

We have repeatedly used this rule for multiplying by a sum.

The product of polynomials. Transformation (simplification) of the product of two polynomials

In general, the product of two polynomials is identically equal to the sum of the product of each term of one polynomial and each term of the other.

Usually use the following rule.

To multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other and add the resulting products.

Abbreviated multiplication formulas. Sum, Difference, and Difference Squares

Some expressions in algebraic transformations have to be dealt with more often than others. Perhaps the most common expressions are \((a + b)^2, \; (a - b)^2 \) and \(a^2 - b^2 \), that is, the square of the sum, the square of the difference, and square difference. You noticed that the names of the indicated expressions seem to be incomplete, so, for example, \((a + b)^2 \) is, of course, not just the square of the sum, but the square of the sum of a and b. However, the square of the sum of a and b is not so common, as a rule, instead of the letters a and b, it contains various, sometimes quite complex expressions.

Expressions \((a + b)^2, \; (a - b)^2 \) are easy to convert (simplify) into polynomials of the standard form, in fact, you have already met with such a task when multiplying polynomials:
\((a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = \)
\(= a^2 + 2ab + b^2 \)

The resulting identities are useful to remember and apply without intermediate calculations. Short verbal formulations help this.

\((a + b)^2 = a^2 + b^2 + 2ab \) - the square of the sum is equal to the sum of the squares and the double product.

\((a - b)^2 = a^2 + b^2 - 2ab \) - the square of the difference is the sum of the squares without doubling the product.

\(a^2 - b^2 = (a - b)(a + b) \) - the difference of squares is equal to the product of the difference and the sum.

These three identities allow in transformations to replace their left parts with right ones and vice versa - right parts with left ones. The most difficult thing in this case is to see the corresponding expressions and understand what the variables a and b are replaced in them. Let's look at a few examples of using abbreviated multiplication formulas.