\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4}{x^2+x+1}\\ \\ \\ \\ \Rightarrow\frac{x^2+x+1+2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Rightarrow x^2+x+1+2x^2-5=4x-4\\ \Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\\ \Leftrightarrow3x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ \)
vậy S=\(\left\{0;1\right\}\)