Bài giải:

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

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

Đáp án đúng là A