Answer
$\dfrac{1}{x^2-2x+4}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{1}{x}+\dfrac{2}{x^2}}{x+\dfrac{8}{x^2}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{x(1)+1(2)}{x^2}}{\dfrac{x^2(x)+1(8)}{x^2}}
\\\\=
\dfrac{\dfrac{x+2}{x^2}}{\dfrac{x^3+8}{x^2}}
\\\\=
\dfrac{x+2}{x^2}\div\dfrac{x^3+8}{x^2}
\\\\=
\dfrac{x+2}{x^2}\cdot\dfrac{x^2}{x^3+8}
\\\\=
\dfrac{x+2}{x^2}\cdot\dfrac{x^2}{(x+2)(x^2-2x+4)}
\\\\=
\dfrac{\cancel{x+2}}{\cancel{x^2}}\cdot\dfrac{\cancel{x^2}}{(\cancel{x+2})(x^2-2x+4)}
\\\\=
\dfrac{1}{x^2-2x+4}
.\end{array}
