Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 4;x\# 9\\
D = \dfrac{{\sqrt x - 7}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{\sqrt x - 7}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x - 7 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 7 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
b)x = 4 + 2\sqrt 3 \left( {tmdk} \right)\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 + 1\\
\Leftrightarrow D = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1 - 3}} = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 2}}\\
= \dfrac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 2} \right)}}{{3 - {2^2}}}\\
= - \left( {3 + 2\sqrt 3 + \sqrt 3 + 2} \right)\\
= - 5 - 3\sqrt 3 \\
c)D < - 1\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 3}} < - 1\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 3}} + 1 < 0\\
\Leftrightarrow \dfrac{{\sqrt x + \sqrt x - 3}}{{\sqrt x - 3}} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 3}}{{\sqrt x - 3}} < 0\\
\Leftrightarrow \dfrac{3}{2} < \sqrt x < 3\\
\Leftrightarrow \dfrac{9}{4} < x < 9\\
Vậy\,\dfrac{9}{4} < x < 9;x\# 4
\end{array}$
