Đáp án:

$\begin{array}{l}
B1)\\
a)Dkxd:a \ge 0;a \ne 1\\
P = \dfrac{{\sqrt a  - 2}}{{1 - \sqrt a }} - \dfrac{{1 + \sqrt a }}{{2 + \sqrt a }} + \dfrac{{3a - 3 + 3\sqrt a }}{{a + \sqrt a  - 2}}\\
 = \dfrac{{ - \left( {\sqrt a  - 2} \right)\left( {\sqrt a  + 2} \right) - \left( {1 + \sqrt a } \right)\left( {\sqrt a  - 1} \right) + 3a - 3 + 3\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 2} \right)}}\\
 = \dfrac{{ - a + 4 - a + 1 + 3a - 3 + 3\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 2} \right)}}\\
 = \dfrac{{a + 3\sqrt a  + 2}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  + 2} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 2} \right)}}\\
 = \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}}\\
b)P = \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}} = \dfrac{{\sqrt a  - 1 + 2}}{{\sqrt a  - 1}} = 1 + \dfrac{2}{{\sqrt a  - 1}}\\
P \in Z\\
 \Rightarrow \dfrac{2}{{\sqrt a  - 1}} \in Z\\
 \Rightarrow \left( {\sqrt a  - 1} \right) \in \left\{ { - 1;1;2} \right\}\left( {do:\sqrt a  - 1 \ge  - 1} \right)\\
 \Rightarrow \sqrt a  \in \left\{ {0;2;3} \right\}\\
 \Rightarrow a \in \left\{ {0;4;9} \right\}\left( {tmdk} \right)\\
B2)\\
a)Dkxd:x \ge 0;x \ne 4;x \ne 9\\
S = \left( {\dfrac{{x - 2\sqrt x }}{{x - 4}} - 1} \right):\left( {\dfrac{{4 - x}}{{x - \sqrt x  - 6}} - \dfrac{{\sqrt x  - 2}}{{3 - \sqrt x }} - \dfrac{{\sqrt x  - 3}}{{\sqrt x  + 2}}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  + 2}} - 1} \right):\dfrac{{4 - x + \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right) - {{\left( {\sqrt x  - 3} \right)}^2}}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 3} \right)}}\\
 = \dfrac{{\sqrt x  - \sqrt x  - 2}}{{\sqrt x  + 2}}.\dfrac{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 3} \right)}}{{4 - x + x - 4 - {{\left( {\sqrt x  - 3} \right)}^2}}}\\
 = \dfrac{{ - 2}}{1}.\dfrac{{\sqrt x  - 3}}{{ - {{\left( {\sqrt x  - 3} \right)}^2}}}\\
 = \dfrac{2}{{\sqrt x  - 3}}\\
b)S = 1\\
 \Rightarrow \dfrac{2}{{\sqrt x  - 3}} = 1\\
 \Rightarrow \sqrt x  - 3 = 2\\
 \Rightarrow \sqrt x  = 5\\
 \Rightarrow x = 25\left( {tmdk} \right)\\
c)S < 0\\
 \Rightarrow \dfrac{2}{{\sqrt x  - 3}} < 0\\
 \Rightarrow \sqrt x  - 3 < 0\\
 \Rightarrow \sqrt x  < 3\\
 \Rightarrow x < 9\\
Vay\,0 \le x < 9;x \ne 4\\
d)S = \dfrac{2}{{\sqrt x  - 3}} \in Z\\
 \Rightarrow \left( {\sqrt x  - 3} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ {1;2;4;5} \right\}\\
 \Rightarrow x \in \left\{ {1;4;16;25} \right\}\\
Do:x \ne 4\\
 \Rightarrow x \in \left\{ {1;16;25} \right\}
\end{array}$