题目内容
已知0<x<4,则
+
的最小值为
.
| 4 |
| x |
| 1 |
| 4-x |
| 9 |
| 4 |
| 9 |
| 4 |
分析:把要求的式子变形为
(
+
)=1+
+
+
,利用基本不等式即可得到
+
的最小值.
| x+(4-x) |
| 4 |
| 4 |
| x |
| 1 |
| 4-x |
| x |
| 4(4-x) |
| 4-x |
| x |
| 1 |
| 4 |
| 4 |
| x |
| 1 |
| 4-x |
解答:解:
+
=
(
+
)=1+
+
+
≥
+2
=
当且仅当
=
时,即x=
时取等号.
故答案为:
| 4 |
| x |
| 1 |
| 4-x |
| x+(4-x) |
| 4 |
| 4 |
| x |
| 1 |
| 4-x |
| x |
| 4(4-x) |
| 4-x |
| x |
| 1 |
| 4 |
| 5 |
| 4 |
|
| 9 |
| 4 |
当且仅当
| x |
| 4(4-x) |
| 4-x |
| x |
| 8 |
| 3 |
故答案为:
| 9 |
| 4 |
点评:本题考查基本不等式的应用,把要求的式子变形为
(
+
)=1+
+
+
,是解题的关键,属于中档题.
| x+(4-x) |
| 4 |
| 4 |
| x |
| 1 |
| 4-x |
| x |
| 4(4-x) |
| 4-x |
| x |
| 1 |
| 4 |
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