Answer
$\dfrac{8-6\sqrt[]{x}+x}{8-2x}$
Work Step by Step
Multiplying by the conjugate of the denominator, the rationalized-denominator form of the given expression, $
\dfrac{4-\sqrt[]{x}}{4+2\sqrt[]{x}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{4-\sqrt[]{x}}{4+2\sqrt[]{x}}\cdot\dfrac{4-2\sqrt[]{x}}{4-2\sqrt[]{x}}
\\\\=
\dfrac{4(4)+4(-2\sqrt[]{x})-\sqrt[]{x}(4)-\sqrt{x}(-2\sqrt[]{x})}{(4)^2-(2\sqrt[]{x})^2}
\\\\=
\dfrac{16-8\sqrt[]{x}-4\sqrt[]{x}+2\sqrt{x(x)}}{16-4x}
\\\\=
\dfrac{16+(-8\sqrt[]{x}-4\sqrt[]{x})+2\sqrt{(x)^2}}{16-4x}
\\\\=
\dfrac{16-12\sqrt[]{x}+2x}{16-4x}
\\\\=
\dfrac{\cancel{2}(8-6\sqrt[]{x}+x)}{\cancel{2}(8-2x)}
\\\\=
\dfrac{8-6\sqrt[]{x}+x}{8-2x}
.\end{array}
Note that all variables are assumed to have positive real numbers.
