Math, asked by PrincessNumera, 1 year ago

Explain the chapter "Algebraic Expression"

With full key points and formulas to solve the sums in that chapter

Cover every topic

Necert - Mathematics > 7th grade

Answers

Answered by Anonymous
22
⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐

✔ ALGEBRIC EXPRESSIONS - An algebric expression is a expression consisting of constant and variable terms connected by one or more of the operations of addition, subtraction, multiplication and division.

✔ An algebric expression having only one term is called a monomial, two terms is called a binomial, three terms is called trinomial, more terms called multinomial.

✔ The terms having same coefficients are called like terms otherwise they are unlike terms.

✔ -5p²q ➡ Literal coefficient :p²q,
Constant coefficient : -5

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FORMULAS :-

1) (a+b)² = a²+2ab+b²
2) (a-b)² = a²-2ab+b²
3) (a+b) (a-b) = a²-b²
4) (a+b+c)² = a²+b²+c²+2(ab+bc+ca)
5) a³+b³ = (a+b) (a²-ab+b²)
6) a³-b³ = (a-b) (a²+ab+b²)
7) a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca)
8) (a+b)²+(a-b)² = 2(a²+b²) [ IT'S A SECRET FORMULA WHICH YOU CAN APPLY IN CERTAIN SITUATIONS. THIS FORMULA IS UNKNOWN TO MAXIMUM MATHEMATICS LEARNERS.]

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✔ I THINK THIS IS ENOUGH FOR YOU. FOR MORE CONCEPT, SURELY ASK ME.

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DO YOU KNOW DEAR ?

✔ a^m × a^n = a^(m+n)
✔ a^m÷a^n = a^(m-n)
✔ (a^m)^n = a^(mn)

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⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐

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Answered by trisha10433
2
hey
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•algebraic expression- combination of constants and variables connected by some or all of the operations +,-,×,÷ is known as algebraic expression

•a symbol having fixed numerical value is called constant

eg - 9,-6,√2 ,π are all constant

• a symbol having different numericals values known as variable

eg-C=2πr (here 2 and π are constant as they have fixed value whileC and r are variables)

terms of algebraic expression
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•the several parts of an algebraic expression seperated by + or - operations

eg - y^6 -64

here this exoresison contains two terms namely y^6 and -64

identities
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•it is an equality which is true for all values of variables


(a+b)^2 =a^2+b^2+2ab

(a-b)^2 = a^2+b^2-2ab

(a+b)^3 = a^3 +b^3+3ab(a+b)

(a-b)^3 = a^3-b^3-3ab(a-b)

a^3+b^3 = (a+b) (a^2-ab+b^2)

a^3-b^3 = (a-b) (a^2+ab+b^2)


hope helped
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