Đáp án:

$\begin{array}{l}
a)x = 9\left( {tmdk} \right)\\
 \Leftrightarrow \sqrt x  = 3\\
B = \dfrac{1}{{\sqrt x  - 1}} = \dfrac{1}{{3 - 1}} = \dfrac{1}{2}\\
b)C = A:B\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{2}{{x - \sqrt x }}} \right):\dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt x .\sqrt x  + 2}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}.\left( {\sqrt x  - 1} \right)\\
 = \dfrac{{x + 2}}{{\sqrt x }}\\
c)C = 3\\
 \Leftrightarrow \dfrac{{x + 2}}{{\sqrt x }} = 3\\
 \Leftrightarrow x + 2 = 3\sqrt x \\
 \Leftrightarrow x - 3\sqrt x  + 2 = 0\\
 \Leftrightarrow \left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = 1 \Leftrightarrow x = 1\left( {ktm} \right)\\
\sqrt x  = 2 \Leftrightarrow x = 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 4\\
d)C - \dfrac{1}{4}\\
 = \dfrac{{x + 2}}{{\sqrt x }} - \dfrac{1}{4}\\
 = \dfrac{{4x + 8 - \sqrt x }}{{4\sqrt x }}\\
 = \dfrac{{4x - \sqrt x  + 8}}{{4\sqrt x }}\\
 = \dfrac{{{{\left( {2\sqrt x } \right)}^2} - 2.2\sqrt x .\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{127}}{{16}}}}{{4\sqrt x }}\\
 = \dfrac{{{{\left( {2\sqrt x  - \dfrac{1}{4}} \right)}^2} + \dfrac{{127}}{{16}}}}{{4\sqrt x }} > 0\\
 \Leftrightarrow C > \dfrac{1}{4}\\
e)C - 2\\
 = \dfrac{{x + 2}}{{\sqrt x }} - 2\\
 = \dfrac{{x + 2 - 2\sqrt x }}{{\sqrt x }}\\
 = \dfrac{{x - 2\sqrt x  + 1 + 1}}{{\sqrt x }}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2} + 1}}{{\sqrt x }} > 0\\
 \Leftrightarrow C > 2\\
f)C = \dfrac{{x + 2}}{{\sqrt x }} = \sqrt x  + \dfrac{2}{{\sqrt x }}\\
 \Leftrightarrow \sqrt x  \in U\left( 2 \right)\\
 \Leftrightarrow \sqrt x  = 2\left( {do:\sqrt x \# 1} \right)\\
 \Leftrightarrow x = 4\left( {tm} \right)\\
Vậy\,x = 4\\
g)C = \dfrac{{x + 2}}{{\sqrt x }} = \sqrt x  + \dfrac{2}{{\sqrt x }}\\
Theo\,Co - si:\\
\sqrt x  + \dfrac{2}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{2}{{\sqrt x }}}  = 2\sqrt 2 \\
C \ge 2\sqrt 2 \\
 \Leftrightarrow GTNN:C = 2\sqrt 2 \,khi:x = 2
\end{array}$