Bài giải:

$A=\frac{1}{{{x}^{2}}-x+1}+1-\frac{{{x}^{2}}+2}{{{x}^{3}}+1}=\frac{x+1+{{x}^{3}}+1-\left( {{x}^{2}}+2 \right)}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\frac{{{x}^{3}}-{{x}^{2}}+x}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\frac{x\left( {{x}^{2}}-x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)}=\frac{x}{x+1}$thay x=99 vào ta có $A=\frac{99}{99+1}=\frac{99}{100}$