a) Điều kiện xác định:$x\ne 1$ $x\ne-\dfrac{1}{2}$
$\begin{gathered} \frac{{4x + 7}}{{x - 1}} = \frac{{12x + 5}}{{8x + 4}} \hfill \\ \Rightarrow (4x + 7)(8x + 4) = (12x + 5)(x - 1) \hfill \\ \Rightarrow 32{x^2} + 72x + 28 = 12{x^2} - 7x - 5 \hfill \\ \Rightarrow 20{x^2} + 79x + 33 = 0 \hfill \\ \Rightarrow \left[ \begin{gathered} x = - \frac{{79 - \sqrt {3601} }}{{40}} \hfill \\ x = - \frac{{79 + \sqrt {3601} }}{{40}} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$ (có lẽ đề nhầm)
b) Điều kiện xác định: $x\ne \pm\dfrac{1}{2}$
$\begin{gathered} \frac{{2x + 1}}{{2x - 1}} - \frac{{2x - 1}}{{2x + 1}} = \frac{8}{{4{x^2} - 1}} \hfill \\ \Rightarrow \frac{{2x + 1}}{{2x - 1}} - \frac{{2x - 1}}{{2x + 1}} = \frac{8}{{(2x - 1)(2x + 1)}} \hfill \\ \Rightarrow {(2x + 1)^2} - {(2x - 1)^2} = 8 \hfill \\ \Rightarrow 8x - 8 = 0 \hfill \\ \Rightarrow x = 1 thỏa mãn \hfill \\ \end{gathered}$
