设函数f(x)=|2x+1|-|x-3| 解不等式f(x)>0 已知关于x的不等式a+3<fx恒成立,求实数a的取值范围

2024-12-30 05:38:45
推荐回答(1个)
回答1:

f(x) = |2x + 1 |- |x - 4| > 0
|2x + 1| > |x - 4|
|2x + 1|² > |x - 4|²
x² + 4x - 5 > 0
x > 1 或 x < - 5

f(x) + 3|x - 4| > m
|2x + 1| + 2|x - 4| > m
x < -1/2 |2x + 1| + 2|x - 4| = -4x + 7 > 9
-1/2 ≤ x ≤ 4 |2x + 1| + 2|x - 4| = 9
x > 4 |2x + 1| + 2|x - 4| = 4x - 7 > 9
m < 9