Đáp án:+Giải thích các bước giải:

 Đk : x $\neq$ 1

$\frac{x+2}{x³-1}$ + $\frac{x}{x²+x+1}$ +$\frac{1}{1-x}$

= $\frac{x+2}{(x-1)(x²+x+1)}$ + $\frac{x(x-1)}{(x-1)(x²+x+1)}$ - $\frac{1(x²+x+1)}{(x-1)(x²+x+1)}$

= $\frac{x+2}{(x-1)(x²+x+1)}$ + $\frac{x²-x}{(x-1)(x²+x+1)}$ - $\frac{x²+x+1}{(x-1)(x²+x+1)}$ 

= $\frac{x+2+x²-x-(x²+x+1)}{(x-1)(x²+x+1)}$

= $\frac{x+2+x²-x-x²-x-1}{(x-1)(x²+x+1)}$

= $\frac{-x+1}{(x-1)(x²+x+1)}$ 

= $\frac{-(x-1)}{(x-1)(x²+x+1)}$

= $\frac{-1}{(x²+x+1)}$