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