Đáp án:
$\begin{array}{l}
B = \left( {\dfrac{{{a^2}}}{{{a^3} - 4a}} - \dfrac{{10}}{{5a + 10}} - \dfrac{7}{{14 - 7a}}} \right):\\
\left( {a + 2 - \dfrac{{{a^2} - 6}}{{a - 2}}} \right)\\
= \left( {\dfrac{{{a^2}}}{{a\left( {{a^2} - 4} \right)}} - \dfrac{{10}}{{5\left( {a + 2} \right)}} - \dfrac{7}{{7\left( {2 - a} \right)}}} \right)\\
:\dfrac{{\left( {a + 2} \right)\left( {a - 2} \right) - {a^2} + 6}}{{a - 2}}\\
= \left( {\dfrac{a}{{{a^2} - 4}} - \dfrac{2}{{a + 2}} + \dfrac{1}{{a - 2}}} \right).\dfrac{{a - 2}}{{{a^2} - 4 - {a^2} + 6}}\\
= \dfrac{{a - 2\left( {a + 2} \right) + a - 2}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}.\dfrac{{a - 2}}{2}\\
= \dfrac{{a - 2a - 4 + a - 2}}{{a + 2}}.\dfrac{1}{2}\\
= \dfrac{{ - 6}}{{a + 2}}.\dfrac{1}{2}\\
= \dfrac{{ - 3}}{{a + 2}}
\end{array}$
