$ĐKXĐ:$ $x \neq 1$
$PT$ $(1) ⇔ \frac{x² +x +1}{(x -1).(x² +x +1)} - \frac{7x²}{(x -1).(x² +x +1)} = \frac{2x.(x -1)}{(x -1).(x² +x +1)}$
$⇒ x² +x +1 - 7x² = 2x.(x -1)$
$⇔ -6x² +x +1 = 2x² -2x$
$⇔ -8x² +3x +1 = 0$
$⇔ -(8x² -3x -1) = 0$
$⇔ -[(2√2x - \frac{3√2}{8})² - \frac{41}{32}] = 0$
$⇔ -[(2√2x - \frac{3√2}{8} - \sqrt{\frac{41}{32}}).((2√2x - \frac{3√2}{8} + \sqrt{\frac{41}{32}}) = 0$
$⇔ \left[ \begin{array}{l}2√2x - \frac{3√2}{8} - \sqrt{\frac{41}{32}}=0\\2√2x - \frac{3√2}{8} + \sqrt{\frac{41}{32}}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\frac{3 + √41}{16}\\x=\frac{3 - √41}{16}\end{array} \right.$ $(T/m$ $đkxđ)$
$Vậy$ $S =$ {$\frac{3 + √41}{16}; \frac{3 - √41}{16}$}
