Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 9\\
C = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
= \dfrac{{\sqrt x \left( {3 - \sqrt x } \right) + x + 9}}{{\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}:\dfrac{{3\sqrt x + 1 - \left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - x + x + 9}}{{\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{2\sqrt x + 4}}\\
= \dfrac{{3\sqrt x + 9}}{{3 + \sqrt x }}.\dfrac{{ - \sqrt x }}{{2\sqrt x + 4}}\\
= \dfrac{{3\left( {\sqrt x + 3} \right)}}{{3 + \sqrt x }}.\dfrac{{ - \sqrt x }}{{2\sqrt x + 4}}\\
= \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}}\\
b)C < - 1\\
\Rightarrow \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}} + 1 < 0\\
\Rightarrow \dfrac{{ - 3\sqrt x + 2\sqrt x + 4}}{{2\sqrt x + 4}} < 0\\
\Rightarrow - \sqrt x + 4 < 0\left( {do:2\sqrt x + 4 > 0} \right)\\
\Rightarrow \sqrt x > 4\\
\Rightarrow x > 16\\
\text{Vậy}\,x > 16
\end{array}$
