若xy-x-y-4=0，求（xy-1）的平方-2x的平方y-2xy的平方+x的平方+y的平方+6xy-2x-2y的值。（要完整的过程）
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解析：
若xy-x-y-4=0，那么：xy=x+y+4，xy-x-y=4
(xy-1)²-2x²y-2xy²+x²+y²+6xy-2x-2y
=(xy-1)²-2xy(x+y)+x²+2xy+y²+4xy-2x-2y
=(xy-1)²-2xy(x+y)+(x+y)²+2(2xy-x-y)
=(xy-1)²-2xy(x+y)+(x+y)²+2(x+y+8)
=(xy-1)²-2xy(x+y)+2(x+y)+(x+y)²+16
=(xy-1)²-2(xy-1)(x+y)+(x+y)²+16
=(xy-1-x-y)²+16
=(4-1)²+16
=9+16
=25
若xy-x-y-4=0，那么：xy=x+y+4，xy-x-y=4
(xy-1)²-2x²y-2xy²+x²+y²+6xy-2x-2y
=(xy-1)²-2xy(x+y)+x²+2xy+y²+4xy-2x-2y
=(xy-1)²-2xy(x+y)+(x+y)²+2(2xy-x-y)
=(xy-1)²-2xy(x+y)+(x+y)²+2(x+y+8)
=(xy-1)²-2xy(x+y)+2(x+y)+(x+y)²+16
=(xy-1)²-2(xy-1)(x+y)+(x+y)²+16
=(xy-1-x-y)²+16
=(4-1)²+16
=9+16
=25
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