Giải thích các bước giải:
$\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}$
$=\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{-\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x^2+4\right)\left(x-2\right)}-\dfrac{-4x^2}{2\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{\left(x^2-2x\right)\left(x-2\right)-\left(-4x^2\right)}{2\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{x^3+4x}{2\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}$
$=\dfrac{x}{2(x-2)}$
