NCTB Class 7 Math Chapter Five Exercise 5.2 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.2 “Algebraic Formulaic and Applications” Exercise 5.2 Solution.
|
Board |
NCTB |
| Class |
7 |
|
Subject |
Math |
| Chapter |
5 |
|
Chapter Name |
“Algebraic Formulaic and Applications” |
| Exercise |
5.2 Solution |
Exercise:- 5.2
(1) Solution:- (4x+3) (4x-3)
= (4x)2 – (3)2
= 16x2 -9
(2) Solution:- (13-12p) (13+12p)
= (13)2 – (12p)2
= 169 – 144p2
(3) Solution:- (ab+3) (ab-3)
= (a b)2 – (3)2
= a2 b2 – 9
(4) Solution:- (10-xy) (10+xy)
= (10)2 – (x y)2
= 100 – x2y2
(5) Solution:- (4x2+3y2) (4x2-3y2)
= (4x2)2 – (3y2)2
= 16x4– 9y4
(6) Solution:- (a-b-c) (a+ b+ c)
= a- (b +c)}{a+ (b +c)}
= {a +(b +c)} {a- (b +c)}
= a2- (b +c)2
= a2– (b2+2bc+c2)
= a2-b2 -2bc-c2
(8) Solution:- (x-1/2a) (x-5/2a)
= x2+ (-1/2a – 5/2a) x + (-1/2 a) (-5/2a)
= x2+ (-a-5a/2)x + 5a2/4
= x2+ (-6a/2)x + 4a2+4
= x2- 3ax + 5/4a2
(9) Solution:- (1/4x – 1/3y) (1/4x + 1/3y)
= (1/4x)2 – (1/3y)2
= x2/16 – y2/9
(10) Solution:- (a4+3a2 x2+ 9x4) (9x4– 3a2 x2 + a4)
= {(a4+9x4) + 3a2 x2} {a4+ 9x4) – 3a2 x2}
= (a4+9x4)2 – (3a2n2)2
= (a4)2+ 2x2a4x9x4+ (9x4)2 – 9a4x4
= a8+ 18a4 x4 + 81x8– 9 a4 x 44
(11) Solution:- (x+1) (x-1) (x2+1)
= (x2-1) (x2+1)
= (x2)2 – (1)2
= x4-1
(12) Solution:- (9a2+b2) (3a+b) (3a-b)
= (9a2+b2) (9a2 – b2)
= (9a2)2 – (b2) 2
= 81a4 – b4
Where is number 7?
Thank YOU vi
Where is 7????
(x²-x+1),(x²+x+1)
=(x²+1-x)(x²+1+x)
=(x²+1)²-x²
=(x²)²+2.x².1+1²-x²
=x⁴+2x²+1-x²
=x⁴+x²+1