Easy x x

Algebra Level 3

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

Simplify the expression above for x > 2 x>2 .

3 2 4 0 1 5

Dorian Loknar by Dorian Loknar , 30, Croatia

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1 solution

Hung Woei Neoh
Apr 22, 2016

For x > 2 x>2 :

∣ x − 2 ∣ = x − 2 ∣ x − 1 ∣ = x − 1 |x-2| = x-2 \quad\quad |x-1| = x-1

Therefore:

x ( x − 2 ) + ∣ x − 2 ∣ ∣ x − 1 ∣ ( x − 2 ) ( x 2 − 1 ) − ( x + 1 ) 2 = x ( x − 2 ) + x − 2 ( x − 1 ) ( x − 2 ) ( x 2 − 1 ) − ( x + 1 ) 2 = ( x − 2 ) ( x + 1 ) ( x − 1 ) ( x − 2 ) ( x 2 − 1 ) − ( x + 1 ) 2 = x + 1 x − 1 ( x + 1 ) ( x − 1 ) − ( x + 1 ) 2 = ( x + 1 ) 2 − ( x + 1 ) 2 = 0 \dfrac{x(x-2)+|x-2|}{|x-1|(x-2)}(x^2-1) - (x+1)^2\\ =\dfrac{x(x-2)+x-2}{(x-1)(x-2)}(x^2-1) - (x+1)^2\\ =\dfrac{(x-2)(x+1)}{(x-1)(x-2)}(x^2-1) - (x+1)^2\\ =\dfrac{x+1}{x-1}(x+1)(x-1) - (x+1)^2\\ =(x+1)^2 - (x+1)^2\\ =\boxed{0}

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