y=x2−x+12x2−2x+3的值域为_.

问题描述:

y=

x2−x+1
2x2−2x+3
的值域为______.

设x2-x+1=t,(x-

1
2
2+
3
4
=t故t≥
3
4

∴y=
x2−x+1
2x2−2x+3
=
t
2t+1
=
1
2+
1
t
1
2

当t=
3
4
1
t
最大,则y最小,最小值y=
1
2+
1
t
=
3
10

故y=
x2−x+1
2x2−2x+3
的值域为[
3
10
1
2
).
故答案为:[
3
10
1
2
).