Đáp án:
$\begin{array}{l}
a)x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow A = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} = \dfrac{{2.3}}{{3 + 3}} = 1\\
b)B = \left( {\dfrac{{15 - \sqrt x }}{{x - 25}} + \dfrac{2}{{\sqrt x + 5}}} \right):\dfrac{{\sqrt x + 3}}{{\sqrt x - 5}}\\
= \dfrac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}.\dfrac{{\sqrt x - 5}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x + 5}}.\dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{1}{{\sqrt x + 3}}\\
c)M = A + B\\
= \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{1}{{\sqrt x + 3}}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{2\sqrt x + 6 - 5}}{{\sqrt x + 3}}\\
= 2 - \dfrac{5}{{\sqrt x + 3}}\\
M \in Z\\
\Leftrightarrow \dfrac{5}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 = 5\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vay\,x = 4\\
d)M = A + B\\
= 2 - \dfrac{5}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{ - 5}}{{\sqrt x + 3}} \ge - \dfrac{5}{3}\\
\Leftrightarrow 2 - \dfrac{5}{{\sqrt x + 3}} \ge 2 - \dfrac{5}{3}\\
\Leftrightarrow M \ge \dfrac{1}{3}\\
\Leftrightarrow GTNN:M = \dfrac{1}{3}\,khi:x = 0
\end{array}$
