Đáp án:
$\begin{array}{l}
c)\dfrac{{x - 2\sqrt x }}{{x - 5\sqrt x + 6}}\left( {x > 0;x \ne 4;x \ne 9} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
d)\dfrac{1}{{\sqrt 2 - 1}} - \dfrac{{3\sqrt 6 - 3\sqrt {10} }}{{\sqrt 3 - \sqrt 5 }} + \dfrac{4}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 2 - 1}}{{2 - 1}} - \dfrac{{3\sqrt 2 \left( {\sqrt 3 - \sqrt 5 } \right)}}{{\sqrt 3 - \sqrt 5 }} + 2\sqrt 2 \\
= \sqrt 2 - 1 - 3\sqrt 2 + 2\sqrt 2 \\
= - 1\\
e)\left( {\dfrac{{\sqrt x }}{{\sqrt x + 2}} - \dfrac{4}{{x + 2\sqrt x }}} \right).\left( {\dfrac{1}{{\sqrt x + 2}} + \dfrac{4}{{x - 4}}} \right)\\
= \dfrac{{\sqrt x .\sqrt x - 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2 + 4}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 2}}{{x - 4}}\\
= \dfrac{1}{{\sqrt x }}
\end{array}$
