Đáp án:
$\begin{array}{l}
\frac{{x + 3}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{4x - 2}}{{{{\left( {2x - 1} \right)}^2}}}\left( {dkxd:x \ne - 1;x \ne \frac{1}{2}} \right)\\
\Leftrightarrow \frac{{x + 3}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{2\left( {2x - 1} \right)}}{{{{\left( {2x - 1} \right)}^2}}}\\
\Leftrightarrow \frac{{x + 3}}{{{{\left( {x + 1} \right)}^2}}} = \frac{2}{{2x - 1}}\\
\Rightarrow \left( {x + 3} \right)\left( {2x - 1} \right) = 2\left( {{x^2} + 2x + 1} \right)\\
\Rightarrow 2{x^2} - x + 6x - 3 = 2{x^2} + 4x + 2\\
\Rightarrow x = 5\left( {tmdk} \right)\\
Vậy\,x = 5
\end{array}$
