Giải phương trình: \(\frac{{ - 7{x^2} + 4}}{{{x^3} + 1}} = \frac{5}{{{x^2} - x + 1}} - \frac{1}{{x + 1}}\)

* Hướng dẫn giải

ĐKXĐ:  \(x \ne  - 1\)

\(\begin{array}{l}\,\,\,\,\,\,\,\,\frac{{ - 7{x^2} + 4}}{{{x^3} + 1}} = \frac{5}{{{x^2} - x + 1}} - \frac{1}{{x + 1}}\\\Leftrightarrow \frac{{ - 7{x^2} + 4}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{5\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} - \frac{{{x^2} - x + 1}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\ \Leftrightarrow \frac{{ - 7{x^2} + 4}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{5\left( {x + 1} \right) - \left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}\\ \Rightarrow  - 7{x^2} + 4 = 5\left( {x + 1} \right) - \left( {{x^2} - x + 1} \right)\\ \Leftrightarrow  - 7{x^2} + 4 = 5x + 5 - {x^2} + x - 1\\ \Leftrightarrow 6{x^2} + 6x = 0\\ \Leftrightarrow 6x\left( {x + 1} \right) = 0\\\Leftrightarrow \left[ \begin{array}{l}x = 0\\x + 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0(tm)\\x =  - 1(ktm)\end{array} \right.\end{array}\)

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