$x^2+x-2x=0$
$⇔x^2-x=0$
$x(x-1)=0$
$⇔x=0$ hoặc $x-1=0$
$⇔x=0$ hoặc $x=1$
Vậy $S=\{0;1\}$
$\frac{1}{x-1}-$ $\frac{3x^2}{x^3-1}=$ $\frac{2x}{x^2+x+1}$ (1) $ĐKXĐ: x$ $\neq1$
$(1)⇔\frac{x^2+x+1}{(x-1)(x^2+x+1)}-$ $\frac{3x^2}{(x-1)(x^2+x+1)}=$ $\frac{2x(x-1)}{(x-1)(x^2+x+1)}$
$⇒x^2+x+1-3x^2=2x(x-1)$
$⇔-2x^2+x+1=2x^2-2x$
$⇔-2x^2-2x^2+x+2x+1=0$
$⇔-4x^2+3x+1=0$
$⇔-4x^2+4x-x+1=0$
$⇔-4x(x-1)-(x-1)=0$
$⇔(x-1)(-4x-1)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\-4x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1(loại)\\x=-1/4)t/m)\end{array} \right.\)
Vậy $S=\{-1/4\}$
