Answer
$-\dfrac{1}{x}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $
\dfrac{2x-x^2}{x^3-8}\div \dfrac{x^2}{x^2+2x+4}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2x-x^2}{x^3-8}\cdot \dfrac{x^2+2x+4}{x^2}
\\\\=
\dfrac{x(2-x)}{(x-2)(x^2+2x+4)}\cdot \dfrac{x^2+2x+4}{x^2}
\\\\=
\dfrac{-x(x-2)}{(x-2)(x^2+2x+4)}\cdot \dfrac{x^2+2x+4}{x^2}
\\\\=
\dfrac{-\cancel{x}(\cancel{x-2})}{(\cancel{x-2})(\cancel{x^2+2x+4})}\cdot \dfrac{\cancel{x^2+2x+4}}{\cancel{x}\cdot x}
\\\\=
\dfrac{-1}{x}
\\\\=
-\dfrac{1}{x}
.\end{array}
