Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne 3;x \ne  - 3\\
B = \dfrac{{24}}{{{x^2} - 9}} - \dfrac{{x - 4}}{{3 - x}} - \dfrac{{x - 1}}{{3 + x}}\\
 = \dfrac{{24 + \left( {x - 4} \right).\left( {x + 3} \right) - \left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{24 + {x^2} - x - 12 - \left( {{x^2} - 4x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{x^2} - x + 12 - {x^2} + 4x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{3x + 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{3\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{3}{{x - 3}}\\
b)\left| {2x + 1} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 5\\
2x + 1 =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 4\\
2x =  - 6
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x =  - 3\left( {ktm} \right)
\end{array} \right.\\
Khi:x = 2\\
 \Leftrightarrow B = \dfrac{3}{{x - 3}} = \dfrac{3}{{2 - 3}} =  - 3\\
c)B =  - \dfrac{3}{5}\\
 \Leftrightarrow \dfrac{3}{{x - 3}} =  - \dfrac{3}{5}\\
 \Leftrightarrow x - 3 =  - 5\\
 \Leftrightarrow x =  - 2\left( {tm} \right)\\
Vay\,x =  - 2\\
d)B < 0\\
 \Leftrightarrow \dfrac{3}{{x - 3}} < 0\\
 \Leftrightarrow x - 3 < 0\\
 \Leftrightarrow x < 3\\
Vay\,x < 3;x \ne  - 3\\
e)B = \dfrac{3}{{x - 3}} \in Z\\
 \Leftrightarrow \left( {x - 3} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Leftrightarrow x \in \left\{ {0;2;4;6} \right\}\left( {tmdk} \right)
\end{array}$