Bài giải:

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

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

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

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

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

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

Đáp án đúng là C