题目
题型:填空题难度:简单来源:不详
1-x |
2x+1 |
答案
|
1 |
2 |
则函数的定义域是[-
1 |
2 |
由柯西不等式得,
y=2
1-x |
2x+1 |
1-x |
2 |
x+
|
4×2 |
1-x+x+
|
当且仅当2
1-x |
2x+1 |
1 |
2 |
则当x=
1 |
2 |
故答案为:3.
核心考点
举一反三
|
A.-2 | B.10 | C.2 | D.-10 |
A.y=-x2 | B.y=
| C.y=(
| D.y=log2x |
1-x |
2x+1 |
|
1 |
2 |
1 |
2 |
1-x |
2x+1 |
1-x |
2 |
x+
|
4×2 |
1-x+x+
|
1-x |
2x+1 |
1 |
2 |
1 |
2 |
|
A.-2 | B.10 | C.2 | D.-10 |
A.y=-x2 | B.y=
| C.y=(
| D.y=log2x |