Question:

Posted by: Anvi M. on 09.05.2020

185x185-115x115 = (185+115) (185-115) = 300x70 = 21000

(0.99)^2 =  (1-0.01)^2 = 1^2 -2x 1x0.01+ 0.01^2 = 1-0.02+0.0001= 0.9801

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From : Rajesh Singh   9820652111

1 (i)    185x185 – 115x115            Using a2 – b2 = (a + b) (a - b)

= 1852 - 1152

= (185 + 115) (185 - 115)

= 300 x 70

= 21000

Ans

(ii)    (0.99)2                                Using (a  – b)2 = a2 – 2ab + b2

= (1 – 0.01)2

= 12 – 2x1x0.01 + 0.012

= 1 – 0.02 + 0.0001

= 1.0001 – 0.02

= 0.9801

Ans

(iii)    7.83 x 7.83 – 1.17 x 1.17        Using a2 – b2 = (a + b) (a - b) in the numerator

6.66

=    7.832 – 1.172

6.66

=    (7.83 + 1.17) (7.83 – 1.17)

6.66

=    9 x 6.66

6.66

=      9

Ans

(i)     ∵   x + 1/x  =  √5

Squaring both the sides

\ (x + 1/x)2 = (√5)2               (a + b)2 = a2 + 2ab + b2

\ x2 + 2(x)(1/x) + (1/x)2 = 5

\ x2 + 2 + 1/x2 = 5

\ x2 + 1/x2 = 5 – 2

\ x2 + (1/x)2 = 3

Ans

(ii)     ∵   x2 + (1/x)2 = 3     (from i above)

Squaring both the sides

\ (x2 + 1/x2)2 = (3)2               (a + b)2 = a2 + 2ab + b2

\ x4 + 2(x2)(1/x2) + (1/x2)2 = 9

\ x4 + 2 + 1/x4 = 9

\ x4 + 1/x4 = 9 – 2

\ x4 + 1/x4 = 7

Ans

∵ (3x + 5y)2 =   9x2 + 2(3x)(5y) + (5y)2                (a + b)2 = a2 + 2ab + b2

\ (3x + 5y)2 =   9x2 + 30xy + 25y2

\ (3x + 5y)2 = 181 + 30(-6)        Substituting the values of  9x2 + 30xy + 25y2 = 181

\ (3x + 5y)2 = 181- 180                                                                         and xy = -6

\ (3x + 5y)2 = 1                         Taking square root on both the sides

\   3x + 5y = + 1

Ans

∵ (x - 1/x)2 = x2 – 2(x)(1/x) + (1/x)2                   (a - b)2 = a2 - 2ab + b2

\ (x - 1/x)2 = x2 – 2 + 1/x2                    Substituting the values of  x2 + 1/x2 = 66

\ (x - 1/x)2 = 66 – 2

\ (x - 1/x)2 = 64                               Taking square root on both the sides

\ x - 1/x = + 8

Ans

∵ (a + b + c)2 = a2 + b2 + c2 +2(ab + bc +ca)   Substituting the given values

\ 02 = 16 + 2(ab + bc + ca)

\ 0 - 16 = 2(ab + bc + ca)

\ -16 = (ab + bc + ca)

2

\ -8 = ab + bc + ca

\ ab + bc + ca = -8

Ans

4x2 +y2 + 25z2 + 4xy -10yz – 20zx

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

Comparing with the formula, (a + b + c)2 = a2 + b2 + c2 +2(ab + bc +ca)

a =2x, b = y, c = -5z

= 2x + y + (-5z)        Substituting the given values of   x =4, y = 3, z = 2

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

= 8 + 3 -10

= 11 - 10

= 1

Ans

∵ (x - 1/x)3 = x3 – 1/x3 - 3(x)(1/x)(x – 1/x)     Using   (a – b)3 = a3 – b3 – 3ab (a - b)

\ (x - 1/x)3 = x3 – 1/x3 - 3(x – 1/x)

\ (x - 1/x)3 + 3(x – 1/x) = x3 – 1/x3

\ x3 – 1/x3 = (x - 1/x)3 + 3(x – 1/x)       Substituting the values of  x - 1/x

\ x3 – 1/x3 = (3 + 2√2)3 + 3(3 + 2√2)

\ x3 – 1/x3 = (3 + 2√2){(3 + 2√2)2 + 3}

\ x3 – 1/x3 = (3 + 2√2)[{9 + 2(3)(2√2) + (2√2)2} + 3]

\ x3 – 1/x3 = (3 + 2√2)(9 + 12√2 + 8 + 3)

\ x3 – 1/x3 = (3 + 2√2)(20 + 12√2)

\ x3 – 1/x3 = (60 +36√2 + 40√2 + 48)

\ x3 – 1/x3 = 108 +76√2

Ans

64x3 – 125z3                    Using      (a – b)3 = a3 – b3 – 3ab (a - b)

= (4x)3 – (5z)3                                                  \ a3 – b3 = (a – b)3 + 3ab (a - b)

= (4x – 5z)3 + 3(4x)(5z)(4x - 5z)   Substituting the values of  4x – 5z and  xz

= (4x – 5z)3 + 60xz(4x - 5z)

= 163 + 60 x 12 x 16

= 4096 + 11520

= 15616

Ans

∵ (x + y + z)2 = x2 + y 2 + z2 + 2(xy + yz + zx)      Substituting the given values

\      82  = x2 + y 2 + z2 + 2 x 20

\      64 = x2 + y 2 + z2 + 40

\      x2 + y 2 + z2 = 64 - 40

\      x2 + y 2 + z2 = 24

Now, x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

= (x + y + z)(x2 + y2 + z2 – (xy – yz – zx))

=  8(24 – 20)

= 8 x 4

= 32

Ans

Using    a3 – b3 = (a – b)(a2 + ab + b2)

155 x 155 x 155 – 55 x 55 x 55

155 x 155 + 155 x 55 + 55 x 55

=    .           1553 - 553         .

1552 + 155 x 55 + 552

=    (155 – 55)(1552 + 155 x 55 + 552)

1552 + 155 x 55 + 552

= 155 – 55

= 100

Ans

∵  a3 + b3 + c3 = 3abc if a + b + c = 0

Here considering       x - y = a,      y – z = b,         z – x = c,

Then                  a + b + c = x – y + y – z + z – x

= 0

\ (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y)(y – z)(z – x)

Ans

∵   x + y = - 4     Taking cube on both the sides

(x + y)3 = (- 4)3                     Using   (a – b)3 = a3 + b3 + 3ab(a + b)

\ x3 + y3 + 3xy(x + y) = - 64   substituting the values of x + y

\ x3 + y3 + 3xy(-4) = - 64

\ x3 + y3 - 12xy = - 64

\x3 + y3 – 12xy + 64 = 0

Ans

From : Rajesh Singh   9820652111

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