Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 9\\
1)x = 25\left( {tmdk} \right)\\
\sqrt x = 5\\
A = \dfrac{7}{{\sqrt x + 8}} = \dfrac{7}{{5 + 8}} = \dfrac{7}{{13}}\\
2)B = \dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{{2\sqrt x - 24}}{{x - 9}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 3} \right) + 2\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x + 5\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 8} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x + 8}}{{\sqrt x + 3}}\\
3)P = A.B\\
= \dfrac{7}{{\sqrt x + 8}}.\dfrac{{\sqrt x + 8}}{{\sqrt x + 3}}\\
= \dfrac{7}{{\sqrt x + 3}}\\
P \in Z\\
\Leftrightarrow 7 \vdots \sqrt x + 3\\
\Leftrightarrow \sqrt x + 3 = 7\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16
\end{array}$
