Question

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

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Answer 1

xy+x-y+1=6
xy+1=6-(x-y)
(1-x²)(1-y²)+4xy=48
1-x²-y²+x²y²+4xy=48
x²y²+2xy+1-(x²-2xy+y²)=48
(xy+1)²-(x-y)²=48
[(x-y)-6]²-(x-y)²=48
x-y=-1
xy=6-(x-y)=7
xy-x+y+1=xy-(x-y)+1=7-(-1)+1=9

追问

（1-x^2)(1-y^2)+4xy=1-y^2-x^2+x^2*y^2+4xy=2xy-y^2-x^2+(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

为什么这答案和你不一样啊？
你似乎错了吧。。