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晌盯+2xy+1)-(x²-2xy+y²)=48
(xy+1)²陪举-(x-y)²=48
(xy+x-y+1)(xy-x+y+1)=48
xy+x-y+1=6
所以芦谨碧xy-x+y+1=8

Answer 2

（1-x²)(1-y²)+4xy=1-y²-x²+x²y²+4xy=2xy-y²-x²昌信镇+(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）耐粗坦咐=48xy-x+y+1=8