Question

This is a question from subject maths need the answer want some big brains​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\rm :\longmapsto\:\dfrac{1}{(x + 1)( {x}^{2} + 2x + 2) } = \dfrac{A}{x + 1} + \dfrac{Bx + C}{ {(x + 1)}^{2} + 1}$

Find the value of A + B.

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{1}{(x + 1)( {x}^{2} + 2x + 2) } = \dfrac{A}{x + 1} + \dfrac{Bx + C}{ {(x + 1)}^{2} + 1}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{(x + 1)( {x}^{2} + 2x + 2) } = \dfrac{A}{x + 1} + \dfrac{Bx + C}{ {x}^{2} + 2x + 2}$

$\rm :\longmapsto\:1 = A( {x}^{2} + 2x + 2) + (Bx + C)(x + 1) - - (1)$

On substituting x = - 1, we get

$\rm :\longmapsto\:1 = A( {( - 1)}^{2} + 2( - 1) + 2) + (B( - 1) + C)( - 1 + 1) - - (1)$

$\rm :\longmapsto\:1 = A(1 - 2 + 2) + ( - B + C)(0)$

$\rm :\longmapsto\:1 = A(1) +0$

$\bf\implies \: \boxed{ \bf{ \: A = 1}}$

On substituting x = 0, in equation (1), we get

$\rm :\longmapsto\:1 = A( 0 + 0 + 2) + (B(0) + C)(0 + 1)$

$\rm :\longmapsto\:1 = 2A + C$

$\rm :\longmapsto\:1 = 2 \times 1 + C$

$\rm :\longmapsto\:1 = 2 + C$

$\rm :\longmapsto\:1 - 2 =C$

$\bf\implies \: \boxed{ \bf{ \: C = - 1}}$

On substituting x = 1, in equation (1), we get

$\rm :\longmapsto\:1 = A( 1 + 2+ 2) + (B + C)(1 + 1)$

$\rm :\longmapsto\:1 = A(5) + (B + C)(2)$

$\rm :\longmapsto\:5A + 2B + 2C = 1$

$\rm :\longmapsto\:5(1) + 2B + 2( - 1) = 1$

$\rm :\longmapsto\:5 + 2B - 2 = 1$

$\rm :\longmapsto\:3 + 2B = 1$

$\rm :\longmapsto\:2B = 1 - 3$

$\rm :\longmapsto\:2B = - 2$

$\bf\implies \: \boxed{ \bf{ \: B = - 1}}$

Hence,

$\bf\implies \: \boxed{ \bf{ \: A + B = 1 - 1 = 0}}$

Answer 2

Answer:

$\: \: ⟼(x+1)(x2+2x+2)1=x+1A+(x+1)2+1Bx+C</p><p></p><p>Find the value of A + B.</p><p></p><p>\large\underline{\sf{Solution-}}Solution−</p><p></p><p>\rm :\longmapsto\:\dfrac{1}{(x + 1)( {x}^{2} + 2x + 2) } = \dfrac{A}{x + 1} + \dfrac{Bx + C}{ {(x + 1)}^{2} + 1}:⟼(x+1)(x2+2x+2)1=x+1A+(x+1)2+1Bx+C</p><p></p><p>can be rewritten as</p><p></p><p>\rm :\longmapsto\:\dfrac{1}{(x + 1)( {x}^{2} + 2x + 2) } = \dfrac{A}{x + 1} + \dfrac{Bx + C}{ {x}^{2} + 2x + 2}:⟼(x+1)(x2+2x+2)1=x+1A+x2+2x+2Bx+C</p><p></p><p>\rm :\longmapsto\:1 = A( {x}^{2} + 2x + 2) + (Bx + C)(x + 1) - - (1):⟼1=A(x2+2x+2)+(Bx+C)(x+1)−−(1)</p><p></p><p>On substituting x = - 1, we get</p><p></p><p>\rm :\longmapsto\:1 = A( {( - 1)}^{2} + 2( - 1) + 2) + (B( - 1) + C)( - 1 + 1) - - (1):⟼1=A((−1)2+2(−1)+2)+(B(−1)+C)(−1+1)−−(1)</p><p></p><p>\rm :\longmapsto\:1 = A(1 - 2 + 2) + ( - B + C)(0):⟼1=A(1−2+2)+(−B+C)(0)</p><p></p><p>\rm :\longmapsto\:1 = A(1) +0:⟼1=A(1)+0</p><p></p><p>\bf\implies \: \boxed{ \bf{ \: A = 1}}⟹A=1</p><p></p><p>On substituting x = 0, in equation (1), we get</p><p></p><p>\rm :\longmapsto\:1 = A( 0 + 0 + 2) + (B(0) + C)(0 + 1):⟼1=A(0+0+2)+(B(0)+C)(0+1)</p><p></p><p>\rm :\longmapsto\:1 = 2A + C:⟼1=2A+C</p><p></p><p>\rm :\longmapsto\:1 = 2 \times 1 + C:⟼1=2×1+C</p><p></p><p>\rm :\longmapsto\:1 = 2 + C:⟼1=2+C</p><p></p><p>\rm :\longmapsto\:1 - 2 =C:⟼1−2=C</p><p></p><p>\bf\implies \: \boxed{ \bf{ \: C = - 1}}⟹C=−1</p><p></p><p>On substituting x = 1, in equation (1), we get</p><p></p><p>\rm :\longmapsto\:1 = A( 1 + 2+ 2) + (B + C)(1 + 1):⟼1=A(1+2+2)+(B+C)(1+1)</p><p></p><p>\rm :\longmapsto\:1 = A(5) + (B + C)(2):⟼1=A(5)+(B+C)(2)</p><p></p><p>\rm :\longmapsto\:5A + 2B + 2C = 1:⟼5A+2B+2C=1</p><p></p><p>\rm :\longmapsto\:5(1) + 2B + 2( - 1) = 1:⟼5(1)+2B+2(−1)=1</p><p></p><p>\rm :\longmapsto\:5 + 2B - 2 = 1:⟼5+2B−2=1</p><p></p><p>\rm :\longmapsto\:3 + 2B = 1:⟼3+2B=1</p><p></p><p>\rm :\longmapsto\:2B = 1 - 3:⟼2B=1−3</p><p></p><p>\rm :\longmapsto\:2B = - 2:⟼2B=−2</p><p></p><p>\bf\implies \: \boxed{ \bf{ \: B = - 1}}⟹B=−1</p><p></p><p>Hence,</p><p></p><p>\bf\implies \: \boxed{ \bf{ \: A + B = 1 - 1 = 0}}⟹A+B=1−1=0</p><p></p><p>$