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

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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
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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
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

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
x²+4y²-2x+8y+5=0
x²-2x+1+4y²+8y+4=0
(x-1)²+4(y+1)²=0
∴x=1,y= -1
(x^4 - y^4)/(2x²+xy-y²) ×(2x-y)/(xy-y²) ÷ 〔(x²+y²)/(y)〕²
= (x+y)(x-y)(x²+y²) / [(2x-y)(x+y)] ×(2x-y) / [y(x-y)] ×y²/(x²+y²)²
= y/(x²+y²)
= -1/2
此题本意是如此,考察分式化简.
但是数值有误,因为x=1,y=-1时,2x²+xy-y²=0,分式无意义

x²+4y^2-2x+8y+5=0题抄错了,应是4y^2
(x-1)^2+(2y+2)^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
=0
分明抄错了,你还不承认,你脑残啊!

x²+4y^2-2x+8y+5=0
(x-1)^2+4(y+1)^2=0
则x=1,y=-1
代入可解