NCTB Class 7 Math Chapter Five Exercise 5.4 Solution

NCTB Class 7 Math Chapter Five Exercise 5.4 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.4“Algebraic Formulaic and Applications” Exercise 5.4 Solution.

Board	NCTB
Class	7
Subject	Math
Chapter	Five
Chapter Name	“Algebraic Formulaic and Applications”
Exercise	5.4 Solution

Exercise:- 5.4

(1) Solution:- (a-5)² = a²– 2xax5= (5)²

(b) = a²-10a+25

(2) Solution:- (x +y)²+ 2(x +y) (x-y) + (x-y)²

= (x +y + x-y)²

= (2x)²

= 4x²

(3) Solution:- We know, 4ab = (a +b)² – (a-b)²

=> 4ab = 4²-2²

= 16-4

= 12

=> a b = 12/4 = 3 (a)

(4) Solution:- If a quantity is divisible without remainder another quantity is called – (c) Multiple

(5) Solution:- Least common multiple of a, a², a (a +b) is a² (a +b) (d) Answer.

(6) Solution:- H.C.F of 2a and 3b is 1 (a)

(7) Solution:- If a, b are real numbers – (d) (i), (ii), (iii) are correct.

(8) Solution:- (d) x y (x +y) (x-y)

(9) Solution:- H.C.F if two algebraic expression, (d) x (x-y)

(10) Solution:- The L.C.M is — (d) x y (x +y) (x+2y)

(11) Solution:- L.C.M of 9x²– 25y²  and 15ax-25xy is (d) 5a (9x²-25y²)

(12) Solution:- H.C.F of x³ y⁵ and a²-b² is (d) 1.

(13) Solution:- x-1/x = 0; (d) (i), (ii), (iii) Correct.

(14) Solution:- a+1/a= 4 ; a²-4a+1= (d) 0.

(15) Solution:- (a+5)² = a²+2.a.5+25

= a²+10a+25

(16) Solution:- We know, 4ab= (a +b)² – (a-b)²

=> 4ab = (8)² – (4)²

= 64-16

= 48

=> a b= 48/4

= 12

(17) Solution:- 1^st, 3a² b² c= 3² x a² x a x a xb²xb²xc²

2^nd, 6ab2 c2= 2×3²x a²xb²xb²xc²xc

Therefore, H.C.F= 3a² b²c

(18)Solution:- 1^st, 5ab² x²= 5²xa²xb²xbxXxX

2^nd, 10a² by² = 2×5²xa²xaxaxb²xyxy

H.C.F = 5ab

(19) Solution:- 1^st, 3a² x² = 3²xa²x ax X x

2^nd, 6axy² = 2×3²x a²x X x y² x y

3^rd, 9 ay²= 3 x3²xa² xy² x y

H.C.F = 3a

(20) Solution:- 1^st, 16a³ x⁴ y = 2x2x2x2xa³xX⁴xy

2^nd, 40a² y³ x= 2²x 2x 2x5xa²xy³x X

3^rd, 28a x³= 2²x2 x7 xa²x X³

H.C.F= 2ax

(21) Solution:- 1^st, a²+ab = a (a +b)

2^nd, a²-b²= (a +b) (a-b)

Therefore, a2- b2= (a +b) (a-b)

Therefore, H.C.F = (a +b)

(22) Solution:- 1^st, x³y – xy³

= x y (x²-y)

= x y (x +y) (x-y)

2^nd, (x-y)² = (x-y) (x-y)

Therefore, H.C.F = (x-y)

(23) Solution:- 1^st, x² +7x+12

= x²+4x+3x+12

= x (x+4) +3(x+4)

= (x+4) (x+3)

2nd, x²+9x+20

= x² + 5x+ 4x+20

= x(x+5) +4 (x+5)

= (x+5) (x+4)

(24) Solution:-  1^st, a³-ab²= a (a²-b²)

= a (a +b) (a-b)

2^nd, a⁴+2a³b + a² b²

= a² (a²+2ab+b²)

= a² (a +b)²

= a. a (a +b) (a +b)

Therefore, H.C.F = a (a +b)

(25) Solution:- 1^st, a²-16= (a)² – (4)²

= (a+4) (a-4)

2^nd, 3a +12 = 3 (a+4)

3^rd, a²+ 5a +4

= a²+ 4a + a+ 4

= a (a +4) +1(a +4)

= (a +4) (a +1)

(27) Solution:- 1^st, 6a³ b² c

= 2x 3x a³x b²xc

2^nd, 9a⁴ bd²= 3 x3 xa⁴x b x d²

Therefore, L.C.M = 2x 3x 3xa⁴xb²xc xd²

= 18a⁴ b² cd²

(28) Solution:- 1^st, 5x² y²= 5x x²xy²

2^nd, 10xz³= 2x 5x X xz³

3^rd, 15y³ z⁴= 3x 5 x y³x z4

Therefore, L.C.M= 2x 3 x5x X²x y³xz⁴

= 30x² y³ z⁴

(29) Solution:- 1^st, 2p² xy2= 2x p²x X x y²

2^nd, 3p q² = 3x p x q²

3^rd, 6pqx² = 2 x 3 x p x q x X²

Therefore, L.C.M= 6p² q² x² y²

(30) Solution:- 1^st, (b²-c²)= (b +c) (b-c)

2^nd, (b +c)² = (b +c)

Therefore, L.C.M= (b +c)² (b-c)

(31) Solution:- 1^st, x²=2x

= x( x+2)

2^nd, x²+3x+2

= x(x+2) +1(x+2)

= (x+2) (x-1)

Therefore, L.C.M= x (x+1) (x+2)

(32) Solution:- 1^st, 9x²-25y²

= (3x)² – (5y)²

= (3x +5y) (3x -5y)

2^nd, 15ax-25ay = 5a (3x-5y)

(33) Solution:- 1^st, x²-3x-10

= x²-5x=2x-10

= (x-5) (x+2)

2^nd, x²-10x+25

= x²– 5x-5x+25

= (x-5) -5 (x-5)

= (x-5) (x-5)

= (x-5)²

Therefore, L.C.M= (x-5)² (x+2)

(34) Solution:- 1^st, a²-7a+12

= a²-3a-4a+12

= a (a-3) -4 (a-3)

= (a-3) (a-4)

2^nd, a²+a-20

= a²+5a-4a-20

= a(a+5) -4(a+5)

= (a-4) (a+5)

3^rd, a²+ 2a-15

= a²+5a-3a-15

= (a-5) -3(a+5)

= (a-3) (a+5)

Therefore, L.C.M, (a-3) (a-4) (a+5)

(35) Solution:-  1^st, x²-8x+15

= x²-5x-3x+15

= x (x-5) -3(x-5)

= (x-5) (x-3)

2^nd, x²-25= (x+5) (x-5)

3^rd, x²+2x-25

= x²-3x+5x-15

= x (x-3) +5 (x-3)

= (x-3) (x+5)

Therefore, L.C.M = (x-3) (x2-25)

(36) Solution:- 1^st, (x+5)

2^nd, x²+5x

= x (x+5)

3^rd, x²+7x+10

= (x²+5x+2x+10)

= (x+5) (x+2)

Therefore, L.C.M= x (x+5) (x+2)

(37) (a) Solution:- (a +b)

= 2x-3+2x+5

= (4x+2)

(b) Solution:- a²= (2x-3)²

= (2x)²– 2. 2x.3+ (3)2

= 4x²-12x +9

(c) Solution:- a b

= (2x-3) (2x=50

= 4x²+10x-6x-15

= 4x²+4x-15

If, x= 2;

4x (2)²+ 4×2-15

= 16+8-15

= 24-15

= 9

(38) (a) Solution:- x²+3x-10

= x²+5x-2x-10

= x(x+5)-2(x+5)

= (x+5) (x-2)

(b) Solution:- 1^st, x²-625

= (x)²– (25)²

= (x+25) (x-25)

= (x+25){(x)² – (5)²}

= (x+25) (x+5) (x-5)

2^nd, x²+3x-10

= (x=5) (x-2)

L.C.M = (x+5) (x-2) (x-5) (x+25)

(c) Solution:- We get,

From 1^st expression, (x+5) (x-2)

2^nd, expression (x²+25) (x+5) (x-5)

L.C.M= (x+5) (x-2) (x-5) (x² +25)

(39) Solution:- (a) (3x-2y+z)²

= {(3x-2y) +z)²

= (3x-2y)²+2(3x-2y) z+ z²

= (3x)² – 2.3x.2y+ (2y)²+ 6xz- 4yz +z²

= 9x²+ 4y²+z²– 12xy +6x²– 4yz

(b) Solution:- 1^st, x²-3x-10

= (x+2)(x-3)

2rd, X³+6x²+8x

= x(x+6x+8)

= x(x²+4x+2x+8)

= x{x (x+4) +2 (x+4)}

= x(x+4) (x+2)

Therefore, H.C.F= (x+2)

(c) Solution:- 3^rd, expression, x⁴-5x³-14x²

= x²(x²-5x-14)

= x²(x²-7x+2x-14)

= x²{x(x-7) +2(x-7)}

= x² (x-7) (x+2)