Đáp án: $\frac{1}{x-1}$ - $\frac{3x²}{x³-1}$ = $\frac{2x}{x²+x+1}$
⇔ $\frac{1}{x-1}$ - $\frac{3x²}{(x-1)(x²+x+1)}$ = $\frac{2x}{x²+x+1}$
⇔ $\frac{1.(x²+x+1)}{(x-1)(x²+x+1)}$ - $\frac{3x²}{(x-1)(x²+x+1)}$ = $\frac{2x.(x-1)}{(x-1)(x²+x+1)}$
⇒ x²+x+1-3x² = 2x²-2x
⇔ x²+x-3x²-2x²+2x= -1
⇔ 3x = -1
⇔ x = $\frac{-1}{3}$