Đáp án:
\[\left[ \begin{array}{l}
x = 0\\
x = \frac{{5 \pm \sqrt {37} }}{6}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{2x - 5}}{{{x^2} - 1}} - \frac{1}{{1 - x}} = \frac{4}{{{x^2} + x + 1}}\\
\Leftrightarrow \frac{{2x - 5}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \frac{1}{{x - 1}} = \frac{4}{{{x^2} + x + 1}}\\
\Leftrightarrow \frac{{2x - 5 + x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{4}{{{x^2} + x + 1}}\\
\Leftrightarrow \frac{{3x - 4}}{{{x^2} - 1}} = \frac{4}{{{x^2} + x + 1}}\\
\Leftrightarrow \left( {3x - 4} \right)\left( {{x^2} + x + 1} \right) = 4\left( {{x^2} - 1} \right)\\
\Leftrightarrow 3{x^3} + 3{x^2} + 3x - 4{x^2} - 4x - 4 = 4{x^2} - 4\\
\Leftrightarrow 3{x^3} - 5{x^2} - x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{{5 \pm \sqrt {37} }}{6}
\end{array} \right.
\end{array}\)
