Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 9\\
x = 16\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 4\\
A = \dfrac{{\sqrt x }}{{\sqrt x - 3}} = \dfrac{4}{{4 - 3}} = 4\\
2)M = A - B\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \left( {\dfrac{7}{{\sqrt x + 1}} - \dfrac{{12}}{{\left( {\sqrt x + 1} \right)\left( {3 - \sqrt x } \right)}}} \right)\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{7}{{\sqrt x + 1}} - \dfrac{{12}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - 7\left( {\sqrt x - 3} \right) - 12}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + \sqrt x - 7\sqrt x + 21 - 12}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 6\sqrt x + 9}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
3)M > - 3\\
\Leftrightarrow \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} + 3 > 0\\
\Leftrightarrow \dfrac{{\sqrt x - 3 + 3\sqrt x + 3}}{{\sqrt x + 1}} > 0\\
\Leftrightarrow 4\sqrt x > 0\\
\Leftrightarrow x \ne 0\\
Vậy\,x > 0;x \ne 9
\end{array}$
