问题 填空题
若x2+y2+6x-2y+10=0,则分式
x-2y
x3-4x2y+4xy2
=______.
答案

x2+y2+6x-2y+10=x2+6x+9+y2-2y+1=(x+3)2+(y-1)2=0,

∵(x+3)2≥0,(y-1)2≥0,

∴(x+3)2=0,(y-1)2=0,

∴x=-3,y=1;

x-2y
x3-4x2y+4xy2
=
x-2y
x(x2-4xy+4y2)
=
x-2y
x(x-2y)2
=
1
x(x-2y) 

将x=-3,y=1代入得

1
15

故答案为

1
15

选择题
单项选择题