Bài giải:

$A=\frac{{{x}^{3}}-1}{x+2}.\left( \frac{1}{x-1}-\frac{x+1}{{{x}^{2}}+x+1} \right)$

$=\frac{{{x}^{3}}-1}{x+2}.\frac{1}{x-1}-\frac{{{x}^{3}}-1}{x+2}.\frac{x+1}{{{x}^{2}}+x+1}$

$=\frac{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}{x+2}.\frac{1}{x-1}-\frac{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}{x+2}.\frac{x+1}{{{x}^{2}}+x+1}$

=$\frac{{{x}^{2}}+x+1}{x+2}-\frac{\left( x-1 \right)\left( x+1 \right)}{x+2}$

=$\frac{{{x}^{2}}+x+1}{x+2}-\frac{{{x}^{2}}-1}{x+2}$

=$\frac{x+2}{x+2}=1$

Trả lời:1