Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 4\\
\left( {\dfrac{{\sqrt x - 1}}{{x - 4}} - \dfrac{{\sqrt x + 1}}{{x + 4\sqrt x + 4}}} \right):\dfrac{{x\sqrt x }}{{{{\left( {4 - x} \right)}^2}}}\\
= \left[ {\dfrac{{\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} - \dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}}} \right].\dfrac{{{{\left( {\sqrt x - 2} \right)}^2}{{\left( {\sqrt x + 2} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{{{\left( {\sqrt x + 2} \right)}^2}\left( {\sqrt x - 2} \right)}}.\dfrac{{{{\left( {\sqrt x - 2} \right)}^2}{{\left( {\sqrt x + 2} \right)}^2}}}{{x\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - \left( {x - \sqrt x - 2} \right)}}{1}.\dfrac{{\sqrt x - 2}}{{x\sqrt x }}\\
= \dfrac{{2\sqrt x }}{1}.\dfrac{{\sqrt x - 2}}{{x\sqrt x }}\\
= \dfrac{{2\sqrt x - 4}}{x}
\end{array}$
