Question

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

Answer 1

(1-x^2)(1-y^2)+4xy=48
1-x²-y²+x²y²+4xy=48
(x²y²+2xy+1)-(x²-2xy+y²)=48
(xy+1)²-(x-y)²=48
(xy+x-y+1)(xy-x+y+1)=48
xy+x-y+1=6
所以xy-x+y+1=8

Answer 2

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