Đáp án:
$\begin{array}{l}
\frac{{{x^2} - 5x + 1}}{{2x + 2}} = \frac{{{x^2} - 4x + 1}}{{x + 1}}\left( {dk:x \ne - 1} \right)\\
\Rightarrow \frac{{{x^2} - 5x + 1}}{{2x + 2}} = \frac{{2{x^2} - 8x + 2}}{{2x + 2}}\\
\Rightarrow {x^2} - 5x + 1 = 2{x^2} - 8x + 2\\
\Rightarrow {x^2} - 3x + 1 = 0\\
\Rightarrow {\left( {x - \frac{3}{2}} \right)^2} = \frac{5}{4}\\
\Rightarrow x = \frac{3}{2} \pm \frac{{\sqrt 5 }}{2} = \frac{{3 \pm \sqrt 5 }}{2}
\end{array}$