已知实数x,y满足x2+4xy+4y2-x-2y+1/4=0,则x+2y的值为 _ .

问题描述:

已知实数x,y满足x2+4xy+4y2-x-2y+

1
4
=0,则x+2y的值为 ___ .

已知等式变形得:x2+4xy+4y2-x-2y+

1
4
=(x+2y)2-(x+2y)+
1
4
=0,
即(x+2y-
1
2
2=0,
∴x+2y-
1
2
=0,
则x+2y=
1
2

故答案为:
1
2