Chapter 1 Solution West Bengal Board : Class 8 Mathematics
West Bengal Board Class 8 Math Solution full Exercise 1.1, 1.2 by Experts. Here in this page WB Board Class 8 Student can find Ganit Prabha Class 8 Math Chapter 1 (ΰ¦ͺΰ§ΰ¦°ΰ§ΰ¦¬ΰ¦ͺΰ¦Ύΰ¦ ΰ§ΰ¦° ΰ¦ͺΰ§ΰ¦¨ΰ¦°ΰ¦Ύΰ¦²ΰ§ΰ¦ΰ¦¨ΰ¦Ύ) Exercise 1.1, 1.2 Solution.
Board |
WB Board |
Class |
8 (Eight) |
Subject |
Mathematics |
Book Name |
Ganit Prabha |
Chapter |
1 |
1.1 Solution
1.2 Solution
(1)
(i) (1+ 5n)
(ii) (2+5n)
(iii) (1+4n)
(6) (b)
(i) x (ii)
(x4 β 4x3 + 6x2) (x2)
= x6 β 4x5 + 6x4
(i) Γ· (ii)
= x4 β 4x3 + 6x2/ x2
= x2 β 4x + 6
(i) x (ii)
= (3m2n3 + 40m3 n4 β 5m4 n5) x (10m2 n2)
= 30m4n5 + 400m5n6 β 50m6n7
And,
(i) Γ· (ii)
= 3m2n3 + 40m3n4 β 5m 4n5 /10m2n2
= 3/10n + 4mn2 β Β½ m2n3
(c) (i) x (ii)
= (49l2 β 100m2) x (7l + 10m)
= 343l2 + 490ml2 β 700m2l β 1000m3
(i) Β (ii)
= 49l2 β 100m2/7l + 10m
= 7l β 10m
(d) (i) x (ii)
= (625a4 β 81b4) x (5a +3b)
= 3125a5 + 1875a4b β 405b4a β 243b5
Β (7)
(i) (a-b) + (b-c) + (c-a)
= a β b + b βc + c βa
= 0
(ii) (a+b) (a-b) + (b+c) (b-c) + (c+a) (c-a)
= a2 β b2 + b2 β c2 + c2 β a2
= 0
(iii) x2 x (x/y β y/x) x (y/x + x/y) x y2
= x2 x (x2 βy2 /xy) x (y2 + x2 /xy) x y2
= x2y2 (x2 β y2) (x2 + y2)/ x2y2
= (x2)2 β (y2)2
= x4 β y4
(iv) a (b-c) + b (c-a) + c (a-b)
= ab β ac + bc β ab + ac β bc
= 0
(v)Β x2 (y2 β z2) + y2 (z2 β x2) + z2 (x2 β y2)
= x2y2 β x2z2 + y2z2 β y2x2 + z2x2 β z2y2
= 0
(vi) (x3 + y3) (x3 β y3) + (y3 + z3) (y3 β z3) + (z3 + x3) (z3 β x3)
= (x3)2 β (y3)2 + (y3)2 β (z3)2 + (z3)2 β (x3)2
= x6 β y6 + y6 β x6 + z6 β x6
= 0
(8)
(i) 5x β 2y
= (5x β 2y)2
= (5x)2 β 2.5x.2y + (2y)2
= 25x2 β 20xy + 4y2
(ii) (7 + 2m)
= (7 + 2m)2
= 72 + 2.7.2m + (2m)2
= 49 + 28m + 4m2
(iii) x + y + z
= (x+y+z)2
= {(x+y) +z}2
= (x+y)2 + 2.(x+y) . z + (z)2
= x2 + 2xy + y2 + 2xz + 2yz + z2
= x2 + y2 + z2 +2xy + 2xz + 2yz
(iv) a + b β c- d
= {(a+b) β (c+d)}2
= (a+b)2 β 2.(a+b).(c+d) + (c+d)2
= a2 + 2ab + b2 β 2(ac + ad + bc + bd) + c2 + 2.c.d. + d2
= a2 + 2ab + b2 -2ac + 2ad β 2bc β 2bd + c2 + 2cd + d2
= a2 + b2 +c2 + d2 + 2ab + 2cd β 2ac β 2bc β 2ad β 2bd
(9)
(i) 9x2 + 9/25y2 β 18x/5y
= (3x)2 + (3/5y)2 β 2.3x.3/5y
= (3x β 3/5y)2
(ii) 25m2 β 70mn + 49n2
= (5m)2 β 2.5m.7n + (7n)2
= (5m β 7n)2
(iii) (2a β b)2 + (4a β 2b) (a+b) + (a+b)2
= (2a β b) + 2 (2a β b) (a+b) + (a+b)2
= {(2a β b) + (a+b)}2
=(2a β b + a + b)2
= (3a)2
= 9a2
(iv) p2/q2 + q2/p2 β 2
= (p/q)2 + (q/p)2 β 2. p/q. q/p
= (p/q β q/p)2
(10)
(i) 391 x 409
= (400 β 9) (400 +9)
= (400)2 β (9)2
(ii) (4x + 3y) (2x β 3y)
= 8x2+ 6xy β 12xy β 9y2
= 8x2 β 6xy β 9y2
= 9x2 β x2 β 6xy β 9y2
= 9x2 β {x2 + 2.x.3y + (3y)2}
= (3x)2 β (x+3y)2
(iii) x
= x.1
= (x+1/2)2 β (x-1/2)2
(11)
(i) 225m2 β 100n2
= 25 {9m2 β 4n2}
= 25 {(3m)2 β (2m)2}
= 25 (3m + 2n) (3m β 2n)
(ii) 25x2 β 1/9 y2z2
= (5x)2 β (1/3yz)2
= (5x + 1/3yz) (5x β 1/3yz)
(iii) 7ax2 + 14ax + 7a
= 7a (x2 + 2x + 1)
= 7a (x2 + x + x + 1)
= 7a {x (x+1) + 1 (x+1)}
= 7a (x+1) (x+1)
(iv) 3x2 β 6x2a2 + 3a4
= 3 (x4 β 2x2a2 + a4)
= 3 {(x2)2 β 2.x2.a2 + (a2)2}
= 3 (x2 β a2)2
= 3 (x2 β a2) (x2 β a2)
= 3 (x + a) (x βa) (x+ a) x -a)
(v) 4b2c2 β (b2 + c2 β a2)
= (2bc)2 β (b2 + c2 β a2)2
= (2bc + b2 + c2 β a2) (2bc β b2 -c2 + a2)
= {(b+c)2 β a 2} {a2 β (b2 β 2bc + c2)}
= {(b+c)2 β (a)2} { a2 β (b-c)2}
= (b + c + a) (b + c β a) (a + b β c) (a β b + c)
(vi) 64ax2 β 49a (x β 2y)2
= a (8x)2 β { 7 (x β 2y)}2
= a (8x)2 β (7x β 14y)2
= a (8x + 7x β 14y) (8x β 7x + 14y)
= a (15x β 14y) (x + 14y)
(vii) x2 β 9 β 4xy + 4y2
= (x2 β 4xy + 4y2) β 9
= {(x)2 β 2.x.2y + (2y)2} β (3)2
= (x β 2y)2 β (3)2
= (x β 2y + 3) (x β 2y β 3)
(viii) x2 β 2x β y2 + 2y
= (x2 β y2) β 2x + 2y
= (x+y) (x-y) β 2 (x-y)
= (x-y) (x+y-z)
(ix) 3 + 2a β a2
= 4 β 1 + 2a β a2
= (2)2 β {1-2a + a2}
= (2)2 β (1 β a)2
= (2 +1 β a) (2 -1 +a)
= (3 βa) (1+a)
(x) x4 β 1
= (x2)2 β (1)2
= (x2 + 1) (x2 β 1)
= (x2 + 1) (x + 1) (x β 1)
(xi) a2 β b2 β c2 + 2bc
= a2 β (b2 β c2 + 2bc)
= a2 β (b β c)2
= (a + b β c) (a β b + c)
(xii) ac + bc + a+ b
= c (a+b) + 1 (a+b)
= (c+1) (a+b)
(12)
(i) (xy + pq) (xy β pq)
= (xy)2 β (pq)2
= x2y2 βΒ p2q2
(ii) 49 x 51
= (50 β 1) (50 + 1)
= (50)2 β (1)2
= 2500 β 1
= 2499
(iii) (2x β y + 3z) (2x + y +3z)
= {(2x + 3z) β y} { (2x + 3z) + y}
= (2x + 3z)2 β (y)2
= (2x)2 + 2.2x.3z + (3z)2 β y2
= 4x2 + 12xz + 9z2 β y2
(v) (a-2) (a+2) (a2+4)
= (a)2 β (2)2 (a2 + 4)
= (a2 β 4) (a2 + 4)
= (a)2 β (4)2
= a4 β 16
(vi) (a + b β c) (b + c β a)
= {(b + (a-c)} {b β (a β c)}
= (b)2 β (a βc)2
= b2 β (a2 β 2ac + c2)
= b2 β a2 + 2ac β c2
(13) x + 1/x = 4
8ab (a2 + b2)
= 8 x 3 x 2 (32 + 22)
= 48 (9 +4)
= 48 x 13
= 624 (Proved)
(e) (x + y)
= (x β y)2 + 2xy
= (3)2 + 2.28
= 9 + 56
= 65
(14)
(a) 2(a2 + b2)
= 2 (a)2 + (b)2
= (a+b)2 + (a-b)2
= (a2 + 2ab + b2) + (a2 β 2ab + b2)
(b) 50x2 + 18y2
= 2 (25x2 + 9y2)
= 2 {(5x)2 + (3y)2}
= (5x + 3y)2 + (5x + 3y)2
(c) a2 + b2 + c2 + d2 + 2 (ac β bd)
= a2 + b2 + c2 + d2 + 2ac β 2bd
= (a2 + 2ac + c2) + (b β 2bd + d2)
= (a+c)2 + (b β d)2
β΄ -x = tx
=) tx = -x
=) t = -x/x
= -1
(x)2 + (1/22) βtx
= (x β Β½)2 + 2.x.1/2 β tx
= (x β 1/x)2 + x β tx βββ (2)
a2 = 102
β΄a = 10
b2 = 12
β΄b = 1
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