Đáp án:

$\begin{array}{l}
B2)\\
a)Dkxd:15x - 1 \ge 0\\
 \Rightarrow x \ge \dfrac{1}{{15}}\\
\dfrac{5}{3}\sqrt {15x - 1}  - \sqrt {15x - 1}  - 2 = \dfrac{1}{3}\sqrt {15x - 1} \\
 \Rightarrow \dfrac{5}{3}\sqrt {15x - 1}  - \sqrt {15x - 1}  - \dfrac{1}{3}\sqrt {15x - 1}  = 2\\
 \Rightarrow \dfrac{1}{3}\sqrt {15x - 1}  = 2\\
 \Rightarrow \sqrt {15x - 1}  = 6\\
 \Rightarrow 15x - 1 = 36\\
 \Rightarrow 15x = 37\\
 \Rightarrow x = \dfrac{{37}}{{15}}\left( {tmdk} \right)\\
b)Dkxd:x \ge 3\\
\sqrt {4x - 12}  + \sqrt {9x - 27}  - 4\sqrt {x - 3} \\
 = \sqrt {9 + 4\sqrt 5 }  - \sqrt 5 \\
 \Rightarrow 2\sqrt {x - 3}  + 3\sqrt {x - 3}  - 4\sqrt {x - 3}  = \sqrt {{{\left( {\sqrt 5  + 2} \right)}^2}}  - \sqrt 5 \\
 \Rightarrow \sqrt {x - 3}  = \sqrt 5  + 2 - \sqrt 5 \\
 \Rightarrow \sqrt {x - 3}  = 2\\
 \Rightarrow x - 3 = 4\\
 \Rightarrow x = 7\left( {tmdk} \right)\\
Vay\,x = 7\\
B3)\\
a)A = \dfrac{1}{{2\sqrt 3  - 2}} - \dfrac{1}{{2\sqrt 3  + 2}}\\
 = \dfrac{{2\sqrt 3  + 2 - 2\sqrt 3  + 2}}{{\left( {2\sqrt 3  - 2} \right)\left( {2\sqrt 3  + 2} \right)}}\\
 = \dfrac{4}{{{{\left( {2\sqrt 3 } \right)}^2} - {2^2}}}\\
 = \dfrac{4}{{12 - 4}} = \dfrac{1}{2}\\
B = \dfrac{{\sqrt x }}{{\sqrt x  - 1}} - \dfrac{{2\sqrt x  - 1}}{{x - \sqrt x }}\\
 = \dfrac{{\sqrt x .\sqrt x  - 2\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{x - 2\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x }}\\
b)Dkxd:x > 0;x \ne 1\\
B = \dfrac{2}{5}A\\
 \Rightarrow \dfrac{{\sqrt x  - 1}}{{\sqrt x }} = \dfrac{2}{5}.\dfrac{1}{2} = \dfrac{1}{5}\\
 \Rightarrow \sqrt x  = 5\sqrt x  - 5\\
 \Rightarrow 4\sqrt x  = 5\\
 \Rightarrow \sqrt x  = \dfrac{5}{4}\\
 \Rightarrow x = \dfrac{{25}}{{16}}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{25}}{{16}}
\end{array}$