已知x²+4y-2x+8y+5=0,求 (x的四次方-y的四次方)/(2x²+xy-y²)×(2x-y)/(xy-y²) ÷ 〔(x²+y²)/(y)〕²的值

问题描述:

已知x&sup2+4y-2x+8y+5=0,求 (x的四次方-y的四次方)/(2x&sup2+xy-y&sup2)×(2x-y)/(xy-y&sup2) ÷ 〔(x&sup2+y&sup2)/(y)〕&sup2的值

x^2+4y^2-2x+8y+5=0
(x-1)^2+4(y+1)^2=0
x=1,y=-1
(x的四次方-y的四次方)/(2x²+xy-y²)×(2x-y)/(xy-y²) ÷ 〔(x²+y²)/(y)〕²
=(1^4-(-1)^4)/(2x²+xy-y²)×(2x-y)/(xy-y²) ÷ 〔(x²+y²)/(y)〕²
=0
就对了

x^2+4y^2-2x+8y+5=0吧(x-1)^2+4(y+1)^2=0x=1,y=-1(x的四次方-y的四次方)/(2x²+xy-y²)×(2x-y)/(xy-y²) ÷ 〔(x²+y²)/(y)〕²=(1^4-(-1)^4)/(2x²+xy-y²)×(2x-...