Russianal expressions 2015

Algebra Level 3

The result of evaluating ( x 3 − 1 x + 1 ⋅ x x 3 + 1 ( x + 1 ) 2 − x ( x − 1 ) 2 + x ⋅ ( 1 − 1 x ) ) − 1 2 \displaystyle\left ( \dfrac{\dfrac{x^{3}-1}{x+1}\cdot\dfrac{x}{x^{3}+1}}{\dfrac{(x+1)^{2}-x}{(x-1)^{2}+x}\cdot\left (1-\dfrac{1}{x}\right )} \right )^{\dfrac{-1}{2}} at x = 2015 x = 2015 can be represented in the form a b \displaystyle\dfrac{a}{b} , where a and b only share the factor 1. What is a + b a+b ?


The answer is 4031.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Chew-Seong Cheong
Jan 16, 2015

( x 3 − 1 x + 1 ˙ x x 3 + 1 ( x + 1 ) 2 − x ( x − 1 ) 2 + x ˙ ( 1 − 1 x ) ) − 1 2 = ( ( x − 1 ) ( x 2 + x + 1 ) x ( x + 1 ) ( x + 1 ) ( x 2 − x + 1 ) x 2 + 2 x + 1 − x x 2 − 2 x + 1 + x ˙ x − 1 x ) − 1 2 \displaystyle \left(\frac{\frac{x^3-1}{x+1}\dot{}\frac{x}{x^3+1}}{\frac{(x+1)^2-x}{(x-1)^2+x}\dot{}\left( 1-\frac{1}{x} \right)} \right)^{-\frac{1}{2}} = \left( \frac {\frac{(x-1)(x^2+x+1)x}{(x+1)(x+1)(x^2-x+1)}}{\frac{x^2+2x+1-x}{x^2-2x+1+x}\dot{}\frac{x-1}{x}} \right)^{-\frac{1}{2}}

= ( x ( x − 1 ) ( x 2 + x + 1 ) ( x + 1 ) 2 ( x 2 − x + 1 ) ( x − 1 ) ( x 2 + x + 1 ) x ( x 2 − x + 1 ) ) − 1 2 = ( x ( x − 1 ) ( x 2 + x + 1 ) ( x + 1 ) 2 ( x 2 − x + 1 ) ˙ x ( x 2 − x + 1 ) ( x − 1 ) ( x 2 + x + 1 ) ) − 1 2 \displaystyle = \left( \frac {\frac{x(x-1)(x^2+x+1)}{(x+1)^2(x^2-x+1)}}{\frac{(x-1)(x^2+x+1)}{x(x^2-x+1)}} \right)^{-\frac{1}{2}} = \left( \frac{x(x-1)(x^2+x+1)}{(x+1)^2(x^2-x+1)}\dot{} \frac{x(x^2-x+1)} {(x-1)(x^2+x+1)} \right)^{-\frac{1}{2}}

= ( x 2 ( x + 1 ) 2 ) − 1 2 = ( x + 1 ) 2 x 2 = x + 1 x = 2016 2015 = \left( \dfrac {x^2}{(x+1)^2}\right)^{-\frac{1}{2}} = \sqrt{\dfrac {(x+1)^2}{x^2}}= \dfrac {x+1}{x} = \dfrac {2016}{2015}

⇒ a + b = 2016 + 2015 = 4031 \Rightarrow a + b = 2016+2015 = \boxed{4031}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Well done! You did the exact way I did.

Hobart Pao - 6 years, 4 months ago

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