问题描述: 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. 1个回答 分类:数学 2014-12-13 问题解答: 我来补答 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=9-1=8原式=17分之8 展开全文阅读