Đáp án:
$\begin{array}{l}
a)\frac{{{{\left( {2x + 1} \right)}^2}}}{5} - \frac{{{{\left( {x - 1} \right)}^2}}}{3} = \frac{{7{x^2} - 14x - 5}}{{15}}\\
\Leftrightarrow \frac{{3\left( {4{x^2} + 4x + 1} \right) - 5\left( {{x^2} - 2x + 1} \right)}}{{15}} = \frac{{7{x^2} - 14x - 5}}{{15}}\\
\Leftrightarrow 12{x^2} + 12x + 3 - 5{x^2} + 10x - 5 = 7{x^2} - 14x - 5\\
\Leftrightarrow 36x = - 3\\
\Leftrightarrow x = - \frac{1}{{12}}\\
b)\frac{{{{\left( {x - 2} \right)}^2}}}{3} - \frac{{\left( {2x - 3} \right)\left( {x + 3} \right)}}{8} + \frac{{{{\left( {x - 4} \right)}^2}}}{6} = 0\\
\Leftrightarrow \frac{{8\left( {{x^2} - 4x + 4} \right) - 3\left( {2{x^2} + 3x - 9} \right) + 4\left( {{x^2} - 8x + 16} \right)}}{{24}} = 0\\
\Leftrightarrow 8{x^2} - 32x + 32 - 6{x^2} - 9x + 27 + 4{x^2} - 32x + 64 = 0\\
\Leftrightarrow 6{x^2} - 73x + 123 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 10,14\\
x = 2
\end{array} \right.
\end{array}$
