Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 4\\
Q = \left( {\dfrac{1}{{\sqrt x - 2}} + \dfrac{{5\sqrt x - 4}}{{2\sqrt x - x}}} \right):\left( {\dfrac{{2 + \sqrt x }}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x - 5\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{\left( {2 + \sqrt x } \right)\left( {\sqrt x - 2} \right) - \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 4 - 4\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x - 4 - x}}\\
= \dfrac{{ - 4\left( {\sqrt x + 1} \right)}}{{ - 4}}\\
= \sqrt x + 1\\
b)x = \dfrac{{3 - \sqrt 5 }}{2}\left( {tmdk} \right)\\
= \dfrac{{6 - 2\sqrt 5 }}{4}\\
= {\left( {\dfrac{{\sqrt 5 - 1}}{2}} \right)^2}\\
\Leftrightarrow \sqrt x = \dfrac{{\sqrt 5 - 1}}{2}\\
\Leftrightarrow Q = \sqrt x + 1 = \dfrac{{\sqrt 5 - 1}}{2} + 1 = \dfrac{{\sqrt 5 + 1}}{2}
\end{array}$
