Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 4;x\# 9\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - 3.\dfrac{{\sqrt x - 1}}{{x - 5\sqrt x + 6}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right) - 3\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 - x + 2\sqrt x + 3 - 3\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - \sqrt x + 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 1}}{{\sqrt x - 3}}\\
= \dfrac{1}{{3 - \sqrt x }}\\
b)A < - 1\\
\Leftrightarrow \dfrac{1}{{3 - \sqrt x }} < - 1\\
\Leftrightarrow \dfrac{1}{{3 - \sqrt x }} + 1 < 0\\
\Leftrightarrow \dfrac{{1 + 3 - \sqrt x }}{{3 - \sqrt x }} < 0\\
\Leftrightarrow \dfrac{{4 - \sqrt x }}{{3 - \sqrt x }} < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 4}}{{\sqrt x - 3}} < 0\\
\Leftrightarrow 3 < \sqrt x < 4\\
\Leftrightarrow 9 < x < 16\\
Vậy\,9 < x < 16
\end{array}$
