题目内容
x∈(0,2),则y=
+
的最小值是
.
| 1 |
| x |
| 4 |
| 2-x |
| 9 |
| 2 |
| 9 |
| 2 |
分析:由题意可得,y=
+
=
(
+
)[x+(2-x)]=
[5+
+
],利用基本不等式可求函数的最小值
| 1 |
| x |
| 4 |
| 2-x |
| 1 |
| 2 |
| 1 |
| x |
| 4 |
| 2-x |
| 1 |
| 2 |
| 2-x |
| x |
| 4x |
| 2-x |
解答:解:∵0<x<2
∴0<2-x<2
则y=
+
=
(
+
)[x+(2-x)]
=
[5+
+
]≥
(5+2
)=
当且仅当
=
即x=
时取等号
故答案为:
∴0<2-x<2
则y=
| 1 |
| x |
| 4 |
| 2-x |
| 1 |
| 2 |
| 1 |
| x |
| 4 |
| 2-x |
=
| 1 |
| 2 |
| 2-x |
| x |
| 4x |
| 2-x |
| 1 |
| 2 |
| 4 |
| 9 |
| 2 |
当且仅当
| 2-x |
| x |
| 4x |
| 2-x |
| 2 |
| 3 |
故答案为:
| 9 |
| 2 |
点评:本题主要考查了利用基本不等式求解函数的最值,解题的关键是灵活利用x+(2-x)=2的条件
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