题目
(1)若x>-1,求函数y=
| f(x) |
| g(x) |
(2)若不等式f(x)>ag(x)在x∈[-2,2]上恒成立,求实数a的取值范围.
答案
∴y=
| f(x) |
| g(x) |
| (x+5)(x+2) |
| x+1 |
设x+1=t,∵x>-1,∴t>0
原式化为y=
| (t-1)2+7(t-1)+10 |
| t |
| t2+5t+4 |
| t |
| 4 |
| t |
解析 |
| f(x) |
| g(x) |
| f(x) |
| g(x) |
| (x+5)(x+2) |
| x+1 |
| (t-1)2+7(t-1)+10 |
| t |
| t2+5t+4 |
| t |
| 4 |
| t |
解析 |