Đáp án:

$\begin{array}{l}
B = \dfrac{{x + 2}}{{x + \sqrt x  - 2}} - \dfrac{{\sqrt x }}{{\sqrt x  + 2}}\\
 = \dfrac{{x + 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}} - \dfrac{{\sqrt x }}{{\sqrt x  + 2}}\\
 = \dfrac{{x + 2 - \sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{x + 2 - x + \sqrt x }}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  + 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{1}{{\sqrt x  - 1}}
\end{array}$

$\begin{array}{l}
1))Dkxd:x \ge 0;x \ne 1\\
b)A = \dfrac{8}{3}.B\\
 \Leftrightarrow \dfrac{{4\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} = \dfrac{8}{3}.\dfrac{1}{{\sqrt x  - 1}}\\
 \Leftrightarrow \dfrac{{4\sqrt x }}{{\sqrt x  + 1}} = \dfrac{8}{3}\\
 \Leftrightarrow 8\sqrt x  + 8 = 12\sqrt x \\
 \Leftrightarrow 4\sqrt x  = 8\\
 \Leftrightarrow \sqrt x  = 2\\
 \Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
B2)\\
Dkxd:x \ge 0;x \ne 1\\
a)P = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{3}{{\sqrt x  + 1}} - \dfrac{{6\sqrt x  - 4}}{{x - 1}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right) + 3\left( {\sqrt x  - 1} \right) - 6\sqrt x  + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + 3\sqrt x  - 3 - 6\sqrt x  + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
b)P < \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{1}{2} < 0\\
 \Leftrightarrow \dfrac{{2\sqrt x  - 1 - \sqrt x  - 1}}{{2\left( {\sqrt x  + 1} \right)}} < 0\\
 \Leftrightarrow \sqrt x  - 2 < 0\\
 \Leftrightarrow \sqrt x  < 2\\
 \Leftrightarrow x < 4\\
Vậy\,0 \le x < 4;x \ne 1
\end{array}$