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
问题描述：
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
答
x2以及y2是什么意思？？？？？？？
答
x2+4y2+x2y2-6xy+1=0 变成x2-4xy+4y2+x2y2-2xy+1=0 变(x-2y)2+(xy-1)2=0
得出x-2y=0 xy-1=0 解出x=2y xy=1 x=根号2 y=根号2/2
x4-y4/2x-y 乘 2xy-y2/xy-y2 除以（x2+y2/xy）2乘1/x+y
=(xy)2/x2+y2
=1/(2+1/2)
=2/5
答
(x⁴ - y⁴) / (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)] × x²y² / (x...