$ĐKXĐ:x-1$$\neq0⇒x$ $\neq1$. $x^2+x+1$ luôn khác 0
$\frac{1}{x-1}+$ $\frac{2x^2-5}{x^3-1}=$ $\frac{4}{x^2+x+1}$
⇔$\frac{x^2+x+1+2x^2-5-4(x-1)}{(x-1)(x^2+x+1)}=0$
⇔$x^2+x+1+2x^2-5-4x+4=0$
⇔$3x^3-3x=0$=>$3x(x-1)=0$
⇔\(\left[ \begin{array}{l}3x=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1(loại)\end{array} \right.\)
Vậy $x=0$