Đáp án:
$\frac{x + 1}{x -1}$
Giải thích các bước giải:
$ĐK : x \neq ± 1 ; x \neq \frac{1}{2}$
$(\frac{1}{x-1} - \frac{x}{1 - x³}.\frac{x² + x + 1}{x+1}) : \frac{2x+1}{x² + 2x + 1}$
$=(\frac{1}{x-1}+\frac{x}{(x-1)(x² + x + 1)}.\frac{x² + x + 1}{x+1}).\frac{x² + 2x + 1}{2x+1}$
$=\frac{1}{x-1}+\frac{x}{(x-1)(x+1)}.\frac{(x+1)²}{2x+1}$
$=\frac{x+1+x}{(x-1)(x+1)}.\frac{(x+1)²}{2x+1}$
$= \frac{2x+1}{(x-1)(x+1)}.\frac{(x+1)²}{2x+1}$
$= \frac{x+1}{x-1}$
