Đáp án:

$\begin{array}{l}
1)x = 9\left( {tmdk} \right)\\
 \Leftrightarrow A = \dfrac{{9 + 3}}{{\sqrt 9  - 2}} = \dfrac{{12}}{{3 - 2}} = 12\\
2)B = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}} + \dfrac{{5\sqrt x  - 2}}{{x - 4}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right) + 5\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2 + 5\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  - 2}}\\
3)P = \dfrac{{4B}}{A} = 4.\dfrac{{\sqrt x }}{{\sqrt x  - 2}}:\dfrac{{x + 3}}{{\sqrt x  - 2}}\\
 = \dfrac{{4\sqrt x }}{{\sqrt x  - 2}}.\dfrac{{\sqrt x  - 2}}{{x + 3}}\\
 = \dfrac{{4\sqrt x }}{{x + 3}}\\
 \Leftrightarrow P.x + 3P = 4\sqrt x \\
 \Leftrightarrow P.x - 4\sqrt x  + 3P = 0\\
\Delta ' \ge 0\\
 \Leftrightarrow 4 - P.3P \ge 0\\
 \Leftrightarrow {P^2} \le \dfrac{4}{3}\\
 \Leftrightarrow 0 \le P \le \dfrac{{2\sqrt 3 }}{3}\left( {do:x \ge 0 \Leftrightarrow P \ge 0} \right)\\
Khi:P \in Z\\
 \Leftrightarrow \left[ \begin{array}{l}
P = 0\\
P = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x = 0;x = 1
\end{array}$