\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{x^2}{x^3-1}\)
\(\Leftrightarrow\) \(\frac{x^2+x+1}{x^3-1}+\frac{2\left(x-1\right)}{x^3-1}=\frac{x^2}{x^3-1}\)
\(\Rightarrow\) x2 + x + 1 + 2(x - 1) = x2
\(\Leftrightarrow\) x2 + x + 1 + 2x - 2 - x2 = 0
\(\Leftrightarrow\) 3x - 1 = 0
\(\Leftrightarrow\) x = \(\frac{1}{3}\)
Vậy S = {\(\frac{1}{3}\)}
Chúc bn học tốt!!