Đáp án:
$\begin{array}{l}
a)A = \dfrac{{\sqrt x - 3}}{{x - 9}} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}\\
= \dfrac{1}{{\sqrt x + 3}} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 3 - 4}}{{\sqrt x + 3}}\\
= 1 - \dfrac{4}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{4}{{\sqrt x + 3}} \le \dfrac{4}{3}\\
\Leftrightarrow - \dfrac{4}{{\sqrt x + 3}} \ge \dfrac{{ - 4}}{3}\\
\Leftrightarrow 1 - \dfrac{4}{{\sqrt x + 3}} \ge - \dfrac{1}{3}\\
\Leftrightarrow A \ge - \dfrac{1}{3}\\
\Leftrightarrow GTNN:A = - \dfrac{1}{3}\,khi:x = 0\\
b)\\
A = m\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}} = m\\
\Leftrightarrow \sqrt x - 1 = m\sqrt x + 3m\\
\Leftrightarrow \left( {1 - m} \right).\sqrt x = 3m + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
\sqrt x = \dfrac{{3m + 1}}{{1 - m}} \ge 0;\sqrt x \ne 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
\dfrac{{3m + 1}}{{1 - m}} \ge 0\\
\dfrac{{3m + 1}}{{1 - m}} \ne 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
\dfrac{{3m + 1}}{{m - 1}} \le 0\\
3m + 1 \ne 3 - 3m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
- \dfrac{1}{3} \le m < 1\\
m \ne \dfrac{1}{3}
\end{array} \right.\\
Vậy\, - \dfrac{1}{3} \le m < 1;m \ne \dfrac{1}{3}
\end{array}$
