Đáp án:

$\begin{array}{l}
B2)\\
a)Dkxd:x \ge 0;x \ne 1\\
P = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{3}{{\sqrt x  + 1}} + \dfrac{{6x - 4}}{{1 - x}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right) + 3\left( {\sqrt x  - 1} \right) - 6x + 4}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + 3\sqrt x  - 3 - 6x + 4}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{ - 5x + 4\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{ - \left( {5x - 4\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{ - \left( {5\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{ - \left( {5\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{{ - 5\sqrt x  - 1}}{{\sqrt x  + 1}}\\
b)x = 7 - 4\sqrt 3 \\
 = 4 - 2.2\sqrt 3  + 3\\
 = {\left( {2 - \sqrt 3 } \right)^2}\\
 \Rightarrow \sqrt x  = 2 - \sqrt 3 \\
 \Rightarrow P = \dfrac{{ - 5\sqrt x  - 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{ - 5.\left( {2 - \sqrt 3 } \right) - 1}}{{2 - \sqrt 3  + 1}}\\
 = \dfrac{{ - 11 + 5\sqrt 3 }}{{3 - \sqrt 3 }}\\
 = \dfrac{{2\sqrt 3  - 9}}{3}\\
c)P < \dfrac{1}{2}\\
 \Rightarrow \dfrac{{ - 5\sqrt x  - 1}}{{\sqrt x  + 1}} < \dfrac{1}{2}\\
 \Rightarrow \dfrac{{ - 5\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{1}{2} < 0\\
 \Rightarrow \dfrac{{ - 10\sqrt x  - 2 - \sqrt x  - 1}}{{2\left( {\sqrt x  + 1} \right)}} < 0\\
 \Rightarrow  - 11\sqrt x  - 3 < 0\\
 \Rightarrow \sqrt x  >  - \dfrac{3}{{11}}\left( {tm} \right)\\
\text{Vậy}\,x \ge 0;x \ne 1\\
B3)\\
a)x = \dfrac{{16}}{9} \Rightarrow \sqrt x  = \dfrac{4}{3}\\
 \Rightarrow A = \dfrac{3}{{3\sqrt x  + 1}} = \dfrac{3}{{3.\dfrac{4}{3} + 1}} = \dfrac{3}{5}\\
b)Dkxd:x \ge 0;x \ne \dfrac{1}{9}\\
A = \dfrac{{\sqrt x  - 1}}{{3\sqrt x  - 1}} - \dfrac{1}{{3\sqrt x  + 1}} - \dfrac{{8\sqrt x }}{{1 - 9x}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {3\sqrt x  + 1} \right) - \left( {3\sqrt x  - 1} \right) + 8\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}\\
 = \dfrac{{3x - 2\sqrt x  - 1 - 3\sqrt x  + 1 + 8\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}\\
 = \dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}\\
c)P = A:B\\
 = \dfrac{3}{{3\sqrt x  + 1}}:\dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}
\end{array}$

$\begin{array}{l}
 = \dfrac{3}{{3\sqrt x  + 1}}.\dfrac{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}{{3\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{3\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
P < \dfrac{5}{9}\\
 \Rightarrow \dfrac{{3\sqrt x  - 1}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} - \dfrac{5}{9} < 0\\
 \Rightarrow \dfrac{{27\sqrt x  - 9 - 5\sqrt x \left( {\sqrt x  + 1} \right)}}{{9\sqrt x \left( {\sqrt x  + 1} \right)}} < 0\\
 \Rightarrow 27\sqrt x  - 9 - 5x - 5\sqrt x  < 0\\
 \Rightarrow  - 5x + 22\sqrt x  - 9 < 0\\
 \Rightarrow \sqrt x  > \dfrac{{11 + 2\sqrt {19} }}{5}\\
 \Rightarrow x > 15,5
\end{array}$