An algebra problem by Razif FA

Algebra Level 2

x 2 − 1 x + 1 + ( x + 1 ) 2 − 1 x + 2 + ( x + 2 ) 2 − 1 x + 3 \dfrac{x^2-1}{x+1} + \dfrac{(x+1)^2-1}{x+2} + \dfrac{(x+2)^2-1}{x+3}

Given that x > 0 x>0 , the expression above simplifies to:

3 x 3x ( x + 1 ) 2 (x+1)^2 ( x − 1 ) 2 (x-1)^2 3 ( x − 1 ) 2 3(x-1)^2 3 x − 1 3x-1 1 1 3 ( x + 1 ) 2 3(x+1)^2 3 x + 1 3x+1

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Razif Fa by Razif FA , 23, Indonesia

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1 solution

Kay Xspre
Oct 16, 2015

Simplify and use the perfect square differences gives ( x + 1 ) ( x − 1 ) x + 1 + ( x + 2 ) ( x ) x + 2 + ( x + 3 ) ( x + 1 ) x + 3 = ( x − 1 ) + ( x ) + ( x + 1 ) = 3 x \frac{(x+1)(x-1)}{x+1}+\frac{(x+2)(x)}{x+2}+\frac{(x+3)(x+1)}{x+3} = (x-1)+(x)+(x+1) = 3x

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