Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
x = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow A = \dfrac{{1 + \sqrt x }}{{\sqrt x - 1}} = \dfrac{{1 + 3}}{{3 - 1}} = 2\\
b)Dkxd:x > 0;x \ne 1\\
S = M.A\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x - 1}}.\left( {\dfrac{{x - 2}}{{x + 2\sqrt x }} + \dfrac{1}{{\sqrt x + 2}}} \right)\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x - 1}}.\dfrac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}.\dfrac{{x + \sqrt x - 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
c)S - 1\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }} - 1\\
= \dfrac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x }}\\
= \dfrac{1}{{\sqrt x }} > 0\\
\Leftrightarrow S > 1\\
d)2S = 2\sqrt x + 5\\
\Leftrightarrow 2.\dfrac{{\sqrt x + 1}}{{\sqrt x }} - 2\sqrt x - 5 = 0\\
\Leftrightarrow \dfrac{{2\sqrt x + 2 - 2x - 5\sqrt x }}{{\sqrt x }} = 0\\
\Leftrightarrow - 2x - 3\sqrt x + 2 = 0\\
\Leftrightarrow 2x + 3\sqrt x - 2 = 0\\
\Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) = 0\\
\Leftrightarrow \sqrt x = \dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{1}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{4}
\end{array}$
