Đáp án:
$\begin{array}{l}
1)Dkxd:x > 0;x\# 4\\
x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow A = \dfrac{{\sqrt x - 5}}{{\sqrt x }} = \dfrac{{3 - 5}}{3} = \dfrac{{ - 2}}{3}\\
2)B = \dfrac{3}{{\sqrt x + 2}} + \dfrac{{\sqrt x }}{{2 - \sqrt x }} + \dfrac{{2x - 3\sqrt x + 6}}{{x - 4}}\\
= \dfrac{{3\left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x + 2} \right) + 2x - 3\sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x - 6 - x - 2\sqrt x + 2x - 3\sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
3)P = A.B\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 - 7}}{{\sqrt x + 2}}\\
= 1 - \dfrac{7}{{\sqrt x + 2}}\\
P \in Z\\
\Leftrightarrow \dfrac{7}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 7\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25\left( {tmdk} \right)\\
Vậy\,x = 25\\
4)\left| P \right| > P\\
\Leftrightarrow P < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 5}}{{\sqrt x + 2}} < 0\\
\Leftrightarrow \sqrt x - 5 < 0\\
\Leftrightarrow \sqrt x < 5\\
\Leftrightarrow x < 25\\
Vậy\,0 \le x < 25;x\# 4
\end{array}$
