Giải giúp mình bài 3 ạ

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
{\rm{DK:}}\,\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
P = \dfrac{2}{{\sqrt x  + 3}} - \dfrac{1}{{\sqrt x  - 3}} + \dfrac{{2\sqrt x }}{{x - 9}}\\
 = \dfrac{2}{{\sqrt x  + 3}} - \dfrac{1}{{\sqrt x  - 3}} + \dfrac{{2\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2.\left( {\sqrt x  - 3} \right) - 1.\left( {\sqrt x  + 3} \right) + 2\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{2\sqrt x  - 6 - \sqrt x  - 3 + 2\sqrt x }}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3.\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
b,\\
P \in Z \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} \in Z \Leftrightarrow \sqrt x  + 3 \in \left\{ { \pm 1; \pm 3} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ { - 6;\, - 4;\, - 2;0} \right\}\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  = 0 \Rightarrow x = 0
\end{array}\)