Đáp án:

$\begin{array}{l}
a)Q = \dfrac{{3x - 3\sqrt x  - 3}}{{x + \sqrt x  - 2}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  + 2}} + \dfrac{{\sqrt x  - 2}}{{1 - \sqrt x }}\\
 = \dfrac{{3x - 3\sqrt x  - 3 - \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{3x - 3\sqrt x  - 3 - x + 1 - x + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}}\\
b)x = 4 + 2\sqrt 3 \left( {tmdk} \right)\\
 = {\left( {\sqrt 3  + 1} \right)^2}\\
 \Leftrightarrow \sqrt x  = \sqrt 3  + 1\\
 \Leftrightarrow Q = \dfrac{{\sqrt 3  + 1 - 2}}{{\sqrt 3  + 1 + 2}} = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 3}}\\
 = \dfrac{{\left( {\sqrt 3  - 1} \right)\left( {3 - \sqrt 3 } \right)}}{{9 - 3}}\\
 = \dfrac{{4\sqrt 3  - 6}}{6}\\
 = \dfrac{{2\sqrt 3  - 3}}{3}\\
c)Q = \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} = \dfrac{1}{3}\\
 \Leftrightarrow 3\sqrt x  - 6 = \sqrt x  + 2\\
 \Leftrightarrow 2\sqrt x  = 8\\
 \Leftrightarrow \sqrt x  = 4\\
 \Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16\\
d)Q > \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} > \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{2\sqrt x  - 4 - \sqrt x  - 2}}{{2\left( {\sqrt x  + 2} \right)}} > 0\\
 \Leftrightarrow \sqrt x  - 6 > 0\\
 \Leftrightarrow \sqrt x  > 6\\
 \Leftrightarrow x > 36\\
Vậy\,x > 36\\
e)Q = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} = \dfrac{{\sqrt x  + 2 - 4}}{{\sqrt x  + 2}}\\
 = 1 - \dfrac{4}{{\sqrt x  + 2}}\\
 \Leftrightarrow \left( {\sqrt x  + 2} \right) \in \left\{ {2;4} \right\}\left( {do:\sqrt x  + 2 \ge 2} \right)\\
 \Leftrightarrow \sqrt x  \in \left\{ {0;2} \right\}\\
 \Leftrightarrow x \in \left\{ {0;4} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;4} \right\}\\
f)x \in \left\{ {0;4} \right\}\\
g)Dkxd:x \ge 4\\
Q.\left( {\sqrt x  + 2} \right) - \sqrt x  - 4 = x - 3\sqrt x  + 2\\
\dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}}.\left( {\sqrt x  + 2} \right) - \sqrt {x - 4}  = x - 3\sqrt x  + 2\\
 \Leftrightarrow \sqrt x  - 2 - \sqrt {x - 4}  = x - 3\sqrt x  + 2\\
 \Leftrightarrow \sqrt {x - 4}  + x - 4\sqrt x  + 4 = 0\\
 \Leftrightarrow \sqrt {x - 4}  + {\left( {\sqrt x  - 2} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
\sqrt {x - 4}  \ge 0\\
{\left( {\sqrt x  - 2} \right)^2} \ge 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x - 4}  = 0\\
\sqrt x  - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
x = 4
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = 4
\end{array}$