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