Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 25\\
x = 9\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 3\\
\Rightarrow A = \dfrac{{4\left( {3 + 1} \right)}}{{25 - 9}} = \dfrac{{4.4}}{{16}} = 1\\
2)B = \left( {\dfrac{{15 - \sqrt x }}{{x - 25}} + \dfrac{2}{{\sqrt x + 5}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 5}}\\
= \dfrac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}.\dfrac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \dfrac{{15 - \sqrt x + 2\sqrt x - 10}}{{\sqrt x + 5}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x + 5}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x + 1}}\\
3)P = A.B\\
= \dfrac{{4\left( {\sqrt x + 1} \right)}}{{25 - x}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{4}{{25 - x}} \in Z\\
\Rightarrow 25 - x \in \left\{ { - 4; - 2; - 1;1;2;4} \right\}\\
\Rightarrow x \in \left\{ {29;27;26;24;23;21} \right\}\left( {tmdk} \right)
\end{array}$
