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.

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/27 21:33:42
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.

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.
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.

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.
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

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
请采纳我

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