Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne \frac{1}{2};x \ne - \frac{1}{2}\\
\frac{{8{x^2}}}{{3\left( {1 - 4{x^2}} \right)}} = \frac{{2x}}{{6x - 3}} - \frac{{1 + 8x}}{{4 + 8x}}\\
\Rightarrow \frac{{8{x^2}}}{{3\left( {1 - 2x} \right)\left( {1 + 2x} \right)}} = \frac{{2x}}{{6x - 3}} - \frac{{1 + 8x}}{{4 + 8x}}\\
\Rightarrow \frac{{8{x^2}.4}}{{12\left( {1 - 2x} \right)\left( {1 + 2x} \right)}} = \frac{{2x.4.\left( {1 + 2x} \right) - 3.\left( {1 + 8x} \right).\left( {1 - 2x} \right)}}{{12\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}\\
\Rightarrow 32{x^2} = 16{x^2} + 8x + 48{x^2} - 18x - 3\\
\Rightarrow 32{x^2} - 10x - 3 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{1}{2}\left( {ktm} \right)\\
x = - \frac{3}{{16}}\left( {tm} \right)
\end{array} \right.\\
Vay\,x = - \frac{3}{{16}}\\
b)Dkxd:x \ne - 1;x \ne 1\\
\frac{{x - 1}}{{x + 1}} - \frac{{{x^2} + x - 2}}{{x + 1}} = \frac{{x + 1}}{{x - 1}} - x - 2\\
\Rightarrow \frac{{x - 1 - {x^2} - x + 2}}{{x + 1}} = \frac{{x + 1}}{{x - 1}} - x - 2\\
\Rightarrow \frac{{ - {x^2} + 1}}{{x + 1}} = \frac{{x + 1}}{{x - 1}} - x - 2\\
\Rightarrow - \left( {x - 1} \right) = \frac{{x + 1}}{{x - 1}} - x - 2\\
\Rightarrow \frac{{x + 1}}{{x - 1}} = 3\\
\Rightarrow x + 1 = 3\left( {x - 1} \right)\\
\Rightarrow x = 2\left( {tmdk} \right)
\end{array}$
