已知实数x,y满足方程组x+xy+y=2+3根号2和x2+y2=6.求|x+y+1|的值.
问题描述：
已知实数x,y满足方程组x+xy+y=2+3根号2和x2+y2=6.求|x+y+1|的值.
请尽快!T-T
答
(X+Y+1)^2=X^2+Y^2+1+2XY+2X+2Y =(X^2+Y^2)+2(X+XY+Y)+1 =6+2*(2+3√2)+1
=6+4+6√2+1=11+6√2