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