Bài giải:

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

Thay x=99; y=50$\Rightarrow A=\frac{99-50-1}{99+50+1}=\frac{48}{150}=\frac{8}{25}$

Trả lời:8/25