已知xy+x-y+1=6,(1-x^2)*(1-y^2)+4xy=48,求xy-x+y+1的值

（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=48
xy+x-y+1=6
∴6（xy-x+y+1）=48
xy-x+y+1=8