作业帮已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以（x2+y2/x
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已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以（x2+y2/xy）2乘1/x+y的值
因为x²+4y²+x²y²-6xy+1=0
(x²-4xy+4y²)+(x²y²-2xy+1)=0
(x-2y)²+(xy-1)²=0
所以x-2y=0且xy-1=0
x=2y,xy=1
2y²=1,y²=1/2
x²=4y²=2
原式=[(x^4-y^4)/(2x-y)]×[(2xy-y²)/(xy-y²)]÷[(x²+y²)/(xy)]²×[1/(x+y)]
={[(x²+y²)(x+y)(x-y)]/(2x-y)}×{[y(2x-y)]/[y(x-y)]}×[(xy)²/(x²+y²)²]×[1/(x+y)] 约分
=(xy)²/(x²+y²)
=1²/(2+1/2)
=1/(5/2)
=2/5
(x²-4xy+4y²)+(x²y²-2xy+1)=0
(x-2y)²+(xy-1)²=0
所以x-2y=0且xy-1=0
x=2y,xy=1
2y²=1,y²=1/2
x²=4y²=2
原式=[(x^4-y^4)/(2x-y)]×[(2xy-y²)/(xy-y²)]÷[(x²+y²)/(xy)]²×[1/(x+y)]
={[(x²+y²)(x+y)(x-y)]/(2x-y)}×{[y(2x-y)]/[y(x-y)]}×[(xy)²/(x²+y²)²]×[1/(x+y)] 约分
=(xy)²/(x²+y²)
=1²/(2+1/2)
=1/(5/2)
=2/5
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