ĐKXĐ: $\begin{cases}x-1\ne 0\\x³-1\ne 0\\x²+x+1\ne 0\end{cases}$
$↔x³-1\ne 0$
$↔x³\ne 1$
$↔x\ne 1$
$\dfrac{1}{x-1}+\dfrac{2x²-5}{x³-1}=\dfrac{4}{x²+x+1}$
$↔\dfrac{x²+x+1}{(x-1)(x²+x+1)}+\dfrac{2x²-5}{(x-1)(x²+x+1)}=\dfrac{4(x-1)}{(x-1)(x²+x+1)}$
$↔x²+x+1+2x²-5=4x-4$
$↔3x²+x-4=4x-4$
$↔3x²+x-4x=-4+4$
$↔3x²-3x=0$
$↔3x(x-1)=0$
\(\leftrightarrow\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
\(\leftrightarrow\left[ \begin{array}{l}x=0(tm)\\x=1(ktm)\end{array} \right.\)
Vậy pt có tập nghiệm $S=\{0\}$
