Đáp án:

$\begin{array}{l}
2)a)Dkxd:x > 0\\
S = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
 = \dfrac{{\sqrt x  + 1 + \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
b)S = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }} = \sqrt x  + 1 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\\
\sqrt x  + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}}  = 2\\
 \Leftrightarrow \sqrt x  + \dfrac{1}{{\sqrt x }} + 1 \ge 3\\
 \Leftrightarrow S \ge 3\\
 \Leftrightarrow GTNN:S = 3\,khi:x = 1\\
3)\\
Dkxd:x \ge 0;x \ne 4\\
a)M = \dfrac{{x + 12}}{{x - 4}} + \dfrac{1}{{\sqrt x  + 2}} - \dfrac{4}{{\sqrt x  - 2}}\\
 = \dfrac{{x + 12 + \sqrt x  - 2 - 4\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x + 12 + \sqrt x  - 2 - 4\sqrt x  - 8}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
b)x = 25\left( {tmdk} \right)\\
 \Leftrightarrow \sqrt x  = 5\\
 \Leftrightarrow M = \dfrac{{5 - 1}}{{5 + 2}} = \dfrac{4}{7}\\
c)\dfrac{1}{M} = 1:\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 2}}\\
 = \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  - 1 + 3}}{{\sqrt x  - 1}}\\
 = 1 + \dfrac{3}{{\sqrt x  - 1}}\\
\dfrac{1}{M} \in Z\\
 \Leftrightarrow \dfrac{3}{{\sqrt x  - 1}} \in Z\\
 \Leftrightarrow \sqrt x  - 1 \in \left\{ { - 1;1;3} \right\}\left( {do:\sqrt x  - 1 \ge  - 1} \right)\\
 \Leftrightarrow \sqrt x  \in \left\{ {0;2;4} \right\}\\
 \Leftrightarrow x \in \left\{ {0;4;16} \right\}\left( {x \ne 4} \right)\\
 \Leftrightarrow x \in \left\{ {0;16} \right\}\\
Vậy\,x \in \left\{ {0;16} \right\}
\end{array}$