Đáp án:
$\begin{array}{l}
D = \left( {\dfrac{1}{{\sqrt a - 3}} - \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 3}}{{\sqrt a - 2}} - \dfrac{{\sqrt a + 2}}{{\sqrt a - 3}}} \right)\\
= \dfrac{{\sqrt a - \sqrt a + 3}}{{\sqrt a \left( {\sqrt a - 3} \right)}}:\dfrac{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{3}{{\sqrt a \left( {\sqrt a - 3} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}{{a - 9 - \left( {a - 4} \right)}}\\
= \dfrac{3}{{\sqrt a }}.\dfrac{{\sqrt a - 2}}{{ - 5}}\\
= \dfrac{{3\left( {\sqrt a - 2} \right)}}{{ - 5\sqrt a }}\\
= \dfrac{{3\left( {2 - \sqrt a } \right)}}{{5\sqrt a }}\\
E = \dfrac{{2\sqrt x }}{{x + 2\sqrt x }} + \dfrac{{x - 1}}{{x + 3\sqrt x + 2}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}} + \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{2}{{\sqrt x + 2}} + \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}
\end{array}$
