Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \ne 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
b)\\
P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right)\\
:\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x - 3}} - \dfrac{1}{2}} \right)\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
:\dfrac{{2\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 3} \right)}}{{2\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
.\dfrac{{2\left( {\sqrt x - 3} \right)}}{{2\sqrt x - 2 - \sqrt x + 3}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{2}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right).2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 6}}{{\sqrt x + 3}}\\
c)P = \dfrac{{ - 6}}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 3} \right) \in \left\{ {3;6} \right\}\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x \in \left\{ {0;3} \right\}\\
\Leftrightarrow x \in \left\{ {0;9} \right\}\\
Do:x \ge 0;x \ne 9\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$
