Đáp án:
$\begin{array}{l}
a)Dkxd:a \ne - 2\\
\frac{{12}}{{{a^3} + 8}} - \frac{1}{{2 + a}} = 1\\
\Rightarrow \frac{{12 - \left( {{a^2} - 2a + 4} \right)}}{{\left( {a + 2} \right)\left( {{a^2} - 2a + 4} \right)}} = 1\\
\Rightarrow \frac{{12 - {a^2} + 2a - 4}}{{{a^3} + 8}} = 1\\
\Rightarrow {a^3} + 8 = - {a^2} + 2a + 8\\
\Rightarrow {a^3} + {a^2} - 2a = 0\\
\Rightarrow a\left( {{a^2} + a - 2} \right) = 0\\
\Rightarrow a\left( {a - 1} \right)\left( {a + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 0\left( {tm} \right)\\
a = 1\left( {tm} \right)\\
a = - 2\left( {ktm} \right)
\end{array} \right.\\
Vậy\,a = 0\,hoặc\,a = 1\\
b)Dkxd:a \ne 1;a \ne - 1\\
\frac{{a + 1}}{{a - 1}} - \frac{{a - 1}}{{a + 1}} = 3a\left( {1 - \frac{{a - 1}}{{a + 1}}} \right)\\
\Rightarrow \frac{{{{\left( {a + 1} \right)}^2} - {{\left( {a - 1} \right)}^2}}}{{\left( {a - 1} \right)\left( {a + 1} \right)}} = 3a.\frac{{a + 1 - a + 1}}{{a + 1}}\\
\Rightarrow \frac{{{a^2} + 2a + 1 - {a^2} + 2a - 1}}{{\left( {a - 1} \right)\left( {a + 1} \right)}} = \frac{{3a.2.\left( {a - 1} \right)}}{{\left( {a - 1} \right)\left( {a + 1} \right)}}\\
\Rightarrow 4a = 6{a^2} - 6a\\
\Rightarrow 6{a^2} - 10a = 0\\
\Rightarrow 2a\left( {3a - 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 0\left( {tm} \right)\\
a = \frac{5}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,a = 0\,hoặc\,a = \frac{5}{3}
\end{array}$
