Giải thích các bước giải:

\(\begin{array}{l}
A = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
3 + \sqrt x  \ne 0\\
9 - x \ne 0\\
x - 3\sqrt x  \ne 0\\
\sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 9
\end{array} \right.\\
b,\\
A = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 3} \right)}} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \dfrac{{\sqrt x \left( {3 - \sqrt x } \right) + x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{\left( {3\sqrt x  + 1} \right) - \left( {\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  - x + x + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{2\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{3\sqrt x  + 9}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{3.\left( {\sqrt x  + 3} \right)}}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{ - 3\sqrt x }}{{2\sqrt x  + 4}}
\end{array}\)