Question

已知xy+x-y+1=6,（1-x²）（1-y²）+4xy=48 求xy-x+y+1的值

Answer 1

（1-x2)(1-y2)+4xy=1-y2-x2+x2y2+4xy=2xy-y2-x2+(1+xy)^2=-(x-y)^2+(1+xy)^2=48

(xy+x-y+1)(xy-x+y+1)=(xy+1)^2-(x-y)^2

即48=6(xy-x+y+1)

xy-x+y+1=8