Đáp án:

$\begin{array}{l}
a)Dkxd:x\# 0;x\# 2\\
B = \left( {\dfrac{{x - 4}}{{x\left( {x - 2} \right)}} + \dfrac{2}{{x - 2}}} \right):\left( {\dfrac{{x + 2}}{x} - \dfrac{x}{{x - 2}}} \right)\\
 = \dfrac{{x - 4 + 2x}}{{x\left( {x - 2} \right)}}:\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - {x^2}}}{{x\left( {x - 2} \right)}}\\
 = \dfrac{{3x - 4}}{{x\left( {x - 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{{x^2} - 4 - {x^2}}}\\
 = \dfrac{{3x - 4}}{{ - 4}}\\
 = \dfrac{{4 - 3x}}{4}\\
b)x =  - 2\left( {tmdk} \right)\\
 \Leftrightarrow B = \dfrac{{4 - 3.\left( { - 2} \right)}}{4} = \dfrac{5}{2}\\
c)\left| B \right| - 2x = 5\\
 \Leftrightarrow \left| B \right| = 2x + 5\left( {dk:x \ge  - \dfrac{5}{2}} \right)\\
 \Leftrightarrow \left| {\dfrac{{4 - 3x}}{4}} \right| = 2x + 5\\
 \Leftrightarrow \left| {3x - 5} \right| = 8x + 20\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 5 = 8x + 20\\
3x - 5 =  - 8x - 20
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x =  - 25\\
11x =  - 15
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\left( {ktm} \right)\\
x =  - \dfrac{{15}}{{11}}\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x =  - \dfrac{{15}}{{11}}\\
d)\left( {2 - x} \right).B\\
 = \left( {2 - x} \right).\dfrac{{4 - 3x}}{4}\\
 = \dfrac{1}{4}.\left( {3x - 4} \right)\left( {x - 2} \right)\\
 = \dfrac{1}{4}.\left( {3{x^2} - 10x + 8} \right)\\
 = \dfrac{1}{4}.3.\left( {{x^2} - \dfrac{{10}}{3}.x + \dfrac{8}{3}} \right)\\
 = \dfrac{3}{4}.\left( {{x^2} - 2.x.\dfrac{5}{3} + \dfrac{{25}}{9} - \dfrac{1}{9}} \right)\\
 = \dfrac{3}{4}.{\left( {x - \dfrac{5}{3}} \right)^2} - \dfrac{3}{4}.\dfrac{1}{9}\\
 = \dfrac{3}{4}{\left( {x - \dfrac{5}{3}} \right)^2} - \dfrac{1}{{12}} \ge \dfrac{{ - 1}}{{12}}\\
 \Leftrightarrow GTNN = \dfrac{{ - 1}}{{12}}\,khi:x = \dfrac{5}{3}\\
e)B =  - 1\\
 \Leftrightarrow \dfrac{{4 - 3x}}{4} =  - 1\\
 \Leftrightarrow 3x - 4 = 4\\
 \Leftrightarrow 3x = 8\\
 \Leftrightarrow x = \dfrac{8}{3}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{8}{3}\\
g)\left| B \right| + 3 < 2x - 1\\
 \Leftrightarrow \left| B \right| < 2x - 4\left( {x > 2} \right)\\
 \Leftrightarrow \dfrac{{\left| {3x - 4} \right|}}{4} < 2x - 4\\
 \Leftrightarrow \left| {3x - 4} \right| < 8x - 16\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 4 < 8x - 16\\
3x - 4 >  - 8x + 16
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x > \dfrac{{20}}{{11}}
\end{array} \right.\\
Vậy\,x > 3
\end{array}$