Answer
$\dfrac{x^{2}-2x}{x^{2}+2x-8}=\dfrac{x}{x+4}$
Work Step by Step
$\dfrac{x^{2}-2x}{x^{2}+2x-8}$
Take out common factor $x$ from the numerator and factor the denominator, then simplify:
$\dfrac{x^{2}-2x}{x^{2}+2x-8}=\dfrac{x(x-2)}{(x+4)(x-2)}=\dfrac{x}{x+4}$
