Đáp án:

\(\begin{array}{l}
a)x \ne  \pm 2\\
b)\dfrac{{3x - 6}}{{2x}}\\
c)\dfrac{9}{2}\\
d)\left[ \begin{array}{l}
x = 6\\
x =  - 6\\
x = 3\\
x =  - 3\\
x =  - 3\\
x = 1\\
x =  - 1
\end{array} \right.\\
e)x \in \emptyset 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne  \pm 2\\
b)B = \left( {\dfrac{x}{{x - 2}} - \dfrac{{{x^2} + 4}}{{{x^2} - 4}} + \dfrac{1}{{x + 2}}} \right):\left( {\dfrac{1}{{x - 2}} + \dfrac{1}{{x + 2}}} \right)\\
 = \dfrac{{x\left( {x + 2} \right) - {x^2} - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{x + 2 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} + 2x - {x^2} + x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{2x}}\\
 = \dfrac{{3x - 6}}{{2x}}\\
c)\left| {2x - 1} \right| = 3\\
 \to \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 =  - 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x =  - 1
\end{array} \right.\\
Thay:x =  - 1\\
 \to B = \dfrac{{3\left( { - 1} \right) - 6}}{{2\left( { - 1} \right)}} = \dfrac{9}{2}\\
d)2B = \dfrac{{6x - 12}}{{2x}} = \dfrac{{3x - 6}}{x} = 3 - \dfrac{6}{x}\\
B \in Z \to \dfrac{6}{x} \in Z\\
 \to x \in U\left( 6 \right)\\
 \to \left[ \begin{array}{l}
x = 6\\
x =  - 6\\
x = 3\\
x =  - 3\\
x =  - 3\\
x = 2\left( l \right)\\
x =  - 2\left( l \right)\\
x = 1\\
x =  - 1
\end{array} \right.\\
e)B = 0\\
 \to \dfrac{{3x - 6}}{{2x}} = 0\\
 \to 3x - 6 = 0\\
 \to x = 2\left( l \right)\\
 \to x \in \emptyset 
\end{array}\)