Đáp án:
$\begin{array}{l}
a)x = 9\left( {tmdk} \right)\\
A = \frac{{4.\left( {\sqrt 9 + 1} \right)}}{{25 - 9}} = \frac{{4\sqrt 9 + 4}}{{16}} = \frac{{4.3 + 4}}{{16}} = 1\\
b)dkxd:x \ge 0;x \ne 25\\
B = \left( {\frac{{15 - \sqrt x }}{{x - 25}} + \frac{2}{{\sqrt x + 5}}} \right):\frac{{\sqrt x + 1}}{{\sqrt x - 5}}\\
= \left( {\frac{{15 - \sqrt x }}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} + \frac{2}{{\sqrt x + 5}}} \right).\frac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \frac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}.\frac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \frac{{\sqrt x + 5}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{1}{{\sqrt x + 1}}\\
c)P = A.B = \frac{{4\left( {\sqrt x + 1} \right)}}{{25 - x}}.\frac{1}{{\sqrt x + 1}} = \frac{4}{{25 - x}}\\
P\,nguyên\,lớn\,nhất\\
\Rightarrow 25 - x = 1\\
\Rightarrow x = 24\left( {tmdk} \right)
\end{array}$
Vậy x=24 thì P đạt giá trị nguyên lớn nhất là 4
