1、
- |x + 3| + a > 6
|x + 3| < a - 6
a ≤ 6:无解
a > 6:
6 - a < x + 3 < a - 6
3 - a < x < a - 9
2、
2f(x) > g(x)
2|x - 1| > - |x + 3| + a
(1) x ≤ -3:
- 2(x - 1) > (x + 3) + a
a < -1 - 3x
a < 8
(2) -3 < x < 1:
- 2(x - 1) > -(x + 3) + a
a < 5 - x
a < 4
(3) x ≥ 1:
2(x - 1) > - (x + 3) + a
a < 1 + 3x
a < 4
∴a < 4