Question

已知x^2+y^2-6x+2y+10=0.求分式(3x^2-2xy-y^2)/(5x^2-7xy+2y^2)的值。

已知x^2+y^2-6x+2y+10=0.求分式(3x^2-2xy-y^2)/(5x^2-7xy+2y^2)的值。
求高手回答。

Answer 1

已知x^2+y^2-6x+2y+10=0
求分式5x^2-7xy+2y^2分之3x^2-2xy-y^2
5x^2-7xy+2y^2分之3x^2-2xy-y^2
=（5x-2y)(x-y)分之(3x+y)(x-y)
=(5x-2y)分之(3x+y)

x^2+y^2-6x+2y+10=0
(x-3)^2+(y+1)^2=0
x=3, y=-1
代入：5x-2y=15+2=17
3x+y=9-1=8
原式=17分之8

Answer 2

x^2+y^2-6x+2y+10
=0(x^2-6x+9)+(y^2+2y+1)
=0(x-3)^2+(y+1)^2=0
x=3 y=-1
（3x^2-2xy-y^2）/（5x^2-7xy+2y^2）=8/17
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Answer 3

x^2+y^2-6x+2y+10=0. (x-3)^2+(y+1)^2=0 x=3 y=-1
(3x^2-2xy-y^2)/(5x^2-7xy+2y^2)=(3x+y)(x-y)/(5x-2y)(x-y)=(3x+y)/(5x-2y)=8/17