Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 0;x \ne \dfrac{1}{2}\\
\dfrac{3}{{2x}} + \dfrac{{3x - 3}}{{2x - 1}} + \dfrac{{2{x^2} + 1}}{{4{x^2} - 2x}} = 0\\
\Leftrightarrow \dfrac{{3.\left( {2x - 1} \right) + \left( {3x - 3} \right).2x + 2{x^2} + 1}}{{2x\left( {2x - 1} \right)}} = 0\\
\Leftrightarrow 6x - 3 + 6{x^2} - 6x + 2{x^2} + 1 = 0\\
\Leftrightarrow 8{x^2} - 2 = 0\\
\Leftrightarrow {x^2} = \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{1}{2}\left( {ktm} \right);x = - \dfrac{1}{2}\left( {tm} \right)\\
Vậy\,x = - \dfrac{1}{2}\\
b)Dkxd:x \ne 1;x \ne - 1\\
\left( {{x^2} - 1} \right).\left( {\dfrac{4}{{x - 1}} - \dfrac{4}{{x + 1}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} = 1\\
\dfrac{{4\left( {x + 1} \right) - 4\left( {x - 1} \right) - \left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1;x = - 1\left( {ktm} \right)\\
4x + 4 - 4x + 4 - {x^2} + 1 = 0
\end{array} \right.\\
\Leftrightarrow 9 - {x^2} = 0\\
\Leftrightarrow x = 3;x = - 3\left( {tm} \right)\\
Vậy\,x = 3;x = - 3
\end{array}$
