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._数学解答

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.

人气：102 ℃　时间：2019-12-16 10:53:20

解答

已知x^2+y^2-6x+2y+10=0求分式5x^2-7xy+2y^2分之3x^2-2xy-y^25x^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=0x=3,y=-1代入：5x-2y=15+2=173x+y=...