Đáp án:

$\begin{array}{l}
B2)\\
Dkxd:x > 0;x\# 4\\
a)A = \left( {\dfrac{1}{{\sqrt x  + 2}} - \dfrac{1}{{\sqrt x  - 2}}} \right):\dfrac{{\sqrt x }}{{x - 2\sqrt x }}\\
 = \dfrac{{\sqrt x  - 2 - \sqrt x  - 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  - 2} \right)}}{{\sqrt x }}\\
 = \dfrac{{ - 4}}{{\sqrt x  + 2}}\\
b)x = 6 - 2\sqrt 5 \left( {tm} \right)\\
 = {\left( {\sqrt 5  - 1} \right)^2}\\
 \Leftrightarrow \sqrt x  = \sqrt 5  - 1\\
 \Leftrightarrow A = \dfrac{{ - 4}}{{\sqrt x  + 2}} = \dfrac{{ - 4}}{{\sqrt 5  - 1 + 2}}\\
 = \dfrac{{ - 4}}{{\sqrt 5  + 1}} = \dfrac{{ - 4\left( {\sqrt 5  - 1} \right)}}{{5 - 1}} = 1 - \sqrt 5 \\
c)A = \dfrac{{ - 2}}{3}\\
 \Leftrightarrow \dfrac{{ - 4}}{{\sqrt x  + 2}} =  - \dfrac{2}{3}\\
 \Leftrightarrow \sqrt x  + 2 = 6\\
 \Leftrightarrow \sqrt x  = 4\\
 \Leftrightarrow x = 16\left( {tm} \right)\\
Vậy\,x = 16\\
d)A <  - 1\\
 \Leftrightarrow \dfrac{{ - 4}}{{\sqrt x  + 2}} <  - 1\\
 \Leftrightarrow \dfrac{{ - 4 + \sqrt x  + 2}}{{\sqrt x  + 2}} < 0\\
 \Leftrightarrow \sqrt x  - 2 < 0\\
 \Leftrightarrow \sqrt x  < 2\\
 \Leftrightarrow x < 4\\
Vậy\,0 < x < 4\\
e)A = \dfrac{{ - 4}}{{\sqrt x  + 2}}\\
Do:\sqrt x  + 2 > 2\\
 \Leftrightarrow \dfrac{1}{{\sqrt x  + 2}} < \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{ - 4}}{{\sqrt x  + 2}} >  - 2\\
 \Leftrightarrow A >  - 2\\
B3)\\
a)Dk:x \ge 1\\
\sqrt {x - 1}  + \sqrt {9x - 9}  + \sqrt {4x - 4}  = 12\\
 \Leftrightarrow \sqrt {x - 1}  + 3\sqrt {x - 1}  + 2\sqrt {x - 1}  = 12\\
 \Leftrightarrow 6\sqrt {x - 1}  = 12\\
 \Leftrightarrow \sqrt {x - 1}  = 2\\
 \Leftrightarrow x - 1 = 4\\
 \Leftrightarrow x = 5\left( {tm} \right)\\
Vậy\,x = 5\\
b)Dk:x \ge 5\\
\sqrt {{x^2} - 5x}  - \sqrt {x - 5}  = 0\\
 \Leftrightarrow \sqrt x .\sqrt {x - 5}  - \sqrt {x - 5}  = 0\\
 \Leftrightarrow \sqrt {x - 5} \left( {\sqrt x  - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 5}  = 0\\
\sqrt x  = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 5\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 5
\end{array}$

$\begin{array}{l}
c)Dkxd:x \ge  - \frac{3}{2}\\
{x^2} + 4x + 5 = 2\sqrt {2x + 3} \\
 \Leftrightarrow {x^2} + 4x + 5 - 2\sqrt {2x + 3}  = 0\\
 \Leftrightarrow {x^2} + 2x + 1 + 2x + 4 - 2\sqrt {2x + 3}  = 0\\
 \Leftrightarrow \left( {{x^2} + 2x + 1} \right) + 2x + 3 - 2\sqrt {2x + 3}  + 1 = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {\sqrt {2x + 3}  - 1} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 0\\
\sqrt {2x + 3}  = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
2x + 3 = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
x =  - 1
\end{array} \right.\\
Vậy\,x =  - 1
\end{array}$