x^3+\frac{x^3}{(x-1)^3}+\frac{3x^2}{x-1}-2=0(x+\frac{x}{x-1})^3-3\frac{x^2}{x-1}(x+\frac{x}{x-1}+3\frac{x^2}{x-1}-2=0
(\frac{x^2}{x-1})^3-3(\frac{x^2}{x-1})^2+3\frac{x^2}{x-1}-2=0
Đặt \frac{x^2}{x-1}=t
ta có pt: t^3-3t^2+3t-2=0.
Suy ra t rồi suy ra x