Đáp án:

$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{\sqrt x  - 2}}{{x - 1}} - \dfrac{{\sqrt x  + 2}}{{x + 2\sqrt x  + 1}}} \right):\dfrac{2}{{{x^2} - 2x + 1}}\\
 = \left( {\dfrac{{\sqrt x  - 2}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} - \dfrac{{\sqrt x  + 2}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right) - \left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}{{{{\left( {\sqrt x  + 1} \right)}^2}.\left( {\sqrt x  - 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
 = \dfrac{{x - \sqrt x  - 2 - x - \sqrt x  + 2}}{1}.\dfrac{{\sqrt x  - 1}}{2}\\
 = \dfrac{{ - 2\sqrt x }}{2}.\left( {\sqrt x  - 1} \right)\\
 = \sqrt x  - x\\
2)A > 0\\
 \Rightarrow \sqrt x  - x > 0\\
 \Rightarrow \sqrt x .\left( {1 - \sqrt x } \right) > 0\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt x  \ne 0\\
1 - \sqrt x  > 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
\sqrt x  < 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
x < 1
\end{array} \right.\\
Vay\,0 < x < 1\\
3)x = 7 - 4\sqrt 3 \left( {tmdk} \right)\\
 \Rightarrow x = 4 - 2.2.\sqrt 3  + 3\\
 \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2}\\
 \Rightarrow \sqrt x  = 2 - \sqrt 3 \\
 \Rightarrow A = \sqrt x  - x\\
 = 2 - \sqrt 3  - 7 + 4\sqrt 3 \\
 = 3\sqrt 3  - 5\\
4)A = \sqrt x  - x\\
 =  - \left( {x - \sqrt x } \right)\\
 =  - \left( {x - 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
 =  - {\left( {\sqrt x  - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
 \Rightarrow GTLN:A = \dfrac{1}{4}\,khi:\sqrt x  = \dfrac{1}{2}\\
 \Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)
\end{array}$