NCTB Class 7 Math Chapter Five Exercise 5.1 Solution

NCTB Class 7 Math Chapter Five Exercise 5.1 Solution by Math Expert. Bangladesh Board Class 7 Math Solution Chapter 5.1 “Algebraic Formulaic and Applications” Exercise 5.1 Solution.

Board

NCTB
Class

7

Subject

Math
Chapter

5

Chapter Name

“Algebraic Formulaic and Applications”
Exercise

5.1 Solution

Exercise:- 5.1

(1) Solution:- (a+5)2

= a2+2x z x 5+52

= a2+10a+25

(2) Solution:- (5x-7)2

= (5x)2 – 2x5xX7+ (7)2

= 25x2-70x+49

(3) Solution:- (3a-11xy)2

= (3a)2 – 2x3ax11xy+ (11xy)2

= 9a2– 66axy+ 121x2 y2

(4) Solution:- (5a2+9m2)2

= (5a2)2+2x5a2.9m2+ (9m2)2

= 25a4+ 90a2 m2+ 81 m4

(5) Solution:-  (55)2

= (50+5)2

= (50)2+ 2x50x5 +52

= 2500+500+25

= 3025

(6) Solution:-  (990)2

= (1000-10)2

= (1000)2 – 2x1000x10+(10)2

= 1000000-20000+100

= 980100

(7) Solution:- (xy-6y)2

= (x y)2– 2.xy.6y+ (6y)2

= x2 y2– 12xy2 +30y2

(8) Solution:- (a x-by)2

= (a x)2– 2.ax.by + (by)2

= a2 x2– 2axby + b2 y2

(9) Solution:- (97)2

= (100-3)2

=  (100)2– 2.100.3+ 32

= 10000- 600+9

= 9409

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

= {(2x+y)-z}2

= (2x+y)2 – 2. (2x+y).z+ z2

= (2x)2 +  2X2xXy+y2– 4zx-2yz+z2

= 4x2+4xy+y2-4zx-2yz+x2

(11) Solution:- (2a-b+3c)2

= {(2a-b)+ 3c}2

= (2a-b)2+2. (2a-b).3c+(3c)2

= (2a)2– 2.2a.b.+b2+12ac-6bc+9c2

= 4a2-4ab+b2+12ac-6bc+9c2

= 4a2+b2+9c2-4ab+12ac-6bc

(13) Solution:- (x2+y2-z2)2

= {(x2+y2)2 – (z2)2}

= (x2+y2)2– 2. (x2+y2). Z2 + (z2)2

= (x2)2+ 2x2y2+ (y2)2– 2x2 z2 – 2y2 z2+ z4

= x4+y4+z4+2x2 y2+ 2y2 z2– 2z2 x2

(14) Solution:- (3x-2y+z)2

= {(3x-2y) + z}2

= (3x-2y)2+ 2. (3x-2y) . z +z2}

= 9x2-12xy+4y2+6xz-4yz+z2

= 9x2+4y2+z2+12xy+6xz-4yz

(15) Solution:- (b c +ca +a b)2

= {(b c +c a) + a b}2

= (b c+ c a)2 + 2. (b c +c a). a b+ (a b)2}

= (b c)2+2.bc.ca+ (ca)2 + 2ab2c +2a2bc+ a2 b2

=b2c2+ c2a2+a2b2+ 2ac2+2ab2c+2a2bc

(17) Solution:- (2a+1)2 – 4a(2a+1)+ 4a2

= (2a1)2– 2. (2a+1). 2a + (2a)2

= {(2a+1) – 2a}2

= (2a+1-2a)2

= (1)2 = 1

(18) Solution:-  Let, 5a+3b= x ; 4a-3b= y

Given, x2+2xy+y2

= (x +y)2

= (5a+3b+4a-3b)2

= (9a)2

= 81a2

(19) Solution:- Let, 7a+b= x ; 7a-b= y

Therefore, given, x2-2xy+y2

= (x-y)2

= (7a+b-7a+b)2

= (2b)2

= 4b2

(20) Solution:- Let, (2x+3y)= a ; 2x-3y= b

Given, a2+ 2ab+b2

= (a +b)2

= (2x+3y+2x-3y)2

= (4x)2

= 16x2

(21) Solution:- (5x-2) = a ; 5x+7= b

Given, a2+b2- 2ab

= (a-b)2

= (5x-2-5x-7)2

= (-9)2

= 81

(22) Solution:- (3ab-cd)2 + 9 (c d-a b) + 6 (3ab-cd) (c d-a b)

= (3ab-cd)2 + 2. (3ab-cd) x 3 (c d-a b) + [3 (c d- a b)]2

Let, 3ab-cd= x ; 3 (c d-a b) = y

Therefore, x2+ 2xy +y2

= (x +y)2

= {(3ab-cd) +3 (c d-a b)}2

= (3ab-cd+3cd-3ab)2

= (2cd)2

= 4c2d2

(23) Solution:- (2x+5y+3z)2 + (5y+3z-x)2- 2 (5y+3z-x) (2x+5y+3z)

Let, (2x+5y+3z) = a ; 5y+3z-x = b

Therefore, a2+b2-2a b

= (a-b)2

= {2x+5y+3z-5y-3z+x}2

= (3x)2

= 9x2

(24) Solution:- (2a-3b+4c)2 + (2a+3b-4c)2 + 2 (2a-3b+4c) (2a+3b-4c)

Therefore, Let, 2a-3b+4c= x and, (2a+3b-4c) = y

Therefore, x2+ y2+2xy

= (x +y)2

= (2a-3b+4c+2a+3b-4c)2

= (4a)2

= 16a2

(25) Solution:- 25x2+36y2– 60xy ; when x= -4, y= -5

Therefore, 25x2 + 36y2– 60xy

= (5x)2 + 2.5x.6y+ (6y)2

= (5x-6y)2

= {5(-4) -6 (-5)}2

= (-20+30)2

= (10)2 = 100

(26) Solution:- 16a2– 24ab- 9b2

= (4a)2 – 2.4a.3b+ (3b)2

= (4a-3b)2

= (4×7-3×6)2

= (28-18)2

= (10)2

= 100

(27) Solution:- (9x2+30x+25)

= (3x)2+2.3x.5+52

= (3x+5)2

= {3x (-2) + 5}2

= (-6+5)2

= (-1)2

=1

(28) Solution:- 81a3+18ac+c2

= (9a)2+ 2.9a.1 +c2

= (9a+c)2

= ({9×7-67}2

= (63-67)2

= (-4)2

= 16

(29) Solution:- a2+ b2= (a +b)2– 2ab

= (5)2– 2x 12

= 25-24

= 1

(31) Solution:- x+ 1/x = 5 (given)

=> (x+1/x)2= (5)2

=> x2+ 1/x2 = 25-2

= x2+ 1/x2 = 23

Both side square,

(x2+1/x2)2 = (23)2

=> (x2-1/x2)2 + 4.x2. 1/x2= 529

=> (x2– 1/x2)

= 529-4

= 525

(32) Solution:- a +b= 8: a-b- 4

We know that,

4ab= (a +b)2 – (a -b)2

= 82– 42

= 64-16

= 48

Therefore, a b= 12

Updated: December 16, 2020 — 8:47 am

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  1. Where is 33 and 34 ??

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