Đáp án:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x\# 9
\end{array} \right.\\
a)Khi:x = 36\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 6\\
\Leftrightarrow Q = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{6 + 1}}{{6 - 3}} = \dfrac{7}{3}\\
b)P = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
M = \dfrac{P}{Q}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
c)M = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} \ge \dfrac{{ - 3}}{3}\\
\Leftrightarrow M \ge - 1\\
\Leftrightarrow GTNN:M = - 1\,khi:x = 0
\end{array}$
