Đáp án:
$\begin{array}{l}
G = \left( {\dfrac{{3\sqrt x - 1}}{{x - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{1}{{x + \sqrt x }}\\
= \dfrac{{3\sqrt x - 1 - \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\sqrt x \left( {\sqrt x + 1} \right)\\
= \dfrac{{3\sqrt x - 1 - \sqrt x - 1}}{{\sqrt x - 1}}.\sqrt x \\
= \dfrac{{2\sqrt x - 2}}{{\sqrt x - 1}}.\sqrt x \\
= 2\sqrt x \\
M = \dfrac{{x - \sqrt x }}{{x - 9}} + \dfrac{1}{{\sqrt x + 3}} - \dfrac{1}{{\sqrt x - 3}}\\
= \dfrac{{x - \sqrt x + \sqrt x - 3 - \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3 - \sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - \sqrt x - 6}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}}
\end{array}$
