KO SPAM Giải pt (theo cách lớp 8) $\dfrac{1}{x-1}-\dfrac{7x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}$

$ĐKXĐ:$ $x \neq 1$

$PT$ $(1) ⇔ \frac{x² +x +1}{(x -1).(x² +x +1)} – \frac{7x²}{(x -1).(x² +x +1)} = \frac{2x.(x -1)}{(x -1).(x² +x +1)}$

$⇒ x² +x +1 – 7x² = 2x.(x -1)$

$⇔ -6x² +x +1 = 2x² -2x$

$⇔ -8x² +3x +1 = 0$

$⇔ -(8x² -3x -1) = 0$

$⇔ -[(2√2x – \frac{3√2}{8})² – \frac{41}{32}] = 0$

$⇔ -[(2√2x – \frac{3√2}{8} – \sqrt{\frac{41}{32}}).((2√2x – \frac{3√2}{8} + \sqrt{\frac{41}{32}}) = 0$

$⇔ \left[ \begin{array}{l}2√2x – \frac{3√2}{8} – \sqrt{\frac{41}{32}}=0\\2√2x – \frac{3√2}{8} + \sqrt{\frac{41}{32}}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\frac{3 + √41}{16}\\x=\frac{3 – √41}{16}\end{array} \right.$ $(T/m$ $đkxđ)$

$Vậy$ $S =$ {$\frac{3 + √41}{16}; \frac{3 – √41}{16}$}