题目
| a•2x+a-2 |
| 2x+1 |
答案
| a•2x+a-2 |
| 2x+1 |
| a(2x+1)-2 |
| 2x+1 |
| 2 |
| 2x+1 |
要使函数为奇函数,则必有f(-x)=-f(x),
即a-
| 2 |
| 2-x+1 |
| 2 |
| 2x+1 |
则2a=
| 2 |
| 2x+1 |
| 2 |
| 2-x+1 |
| 2 |
| 2x+1 |
| 2•2x |
| 1+2x |
| 2(2x+1) |
| 2x+1 |
即a=1.
故答案为:1
| a•2x+a-2 |
| 2x+1 |
| a•2x+a-2 |
| 2x+1 |
| a(2x+1)-2 |
| 2x+1 |
| 2 |
| 2x+1 |
| 2 |
| 2-x+1 |
| 2 |
| 2x+1 |
| 2 |
| 2x+1 |
| 2 |
| 2-x+1 |
| 2 |
| 2x+1 |
| 2•2x |
| 1+2x |
| 2(2x+1) |
| 2x+1 |