题目
| 1-2x |
| 2x+1 |
答案
| 1-2x |
| 2x+1 |
| 1-2x1 |
| 2x1+1 |
| 1-2x2 |
| 2x2+1 |
=
| (1-2x1)(2x2+1)-(1-2x2)(2x1+1) |
| (2x1+1)(2x2+1) |
| 2•(2x2-2x1) |
| (2x1+1)(2x2+1) |
由题设可得2x2-2x1>0,2x1>0,2x2>0,∴
| 2•(2x2-2x1) |
| (2x1+1)(2x2+1) |
即f(x1)-f(x2),故有f(x1)>f(x2),故 f(x)在(-∞,+∞)上为减函数.
| 1-2x |
| 2x+1 |
| 1-2x |
| 2x+1 |
| 1-2x1 |
| 2x1+1 |
| 1-2x2 |
| 2x2+1 |
| (1-2x1)(2x2+1)-(1-2x2)(2x1+1) |
| (2x1+1)(2x2+1) |
| 2•(2x2-2x1) |
| (2x1+1)(2x2+1) |
| 2•(2x2-2x1) |
| (2x1+1)(2x2+1) |