第2个回答 2011-08-20
分析:要对k分开讨论
解: (i) k < 0, 则方程无解. (绝对值一定为非负数)
(ii) k = 0, 则方程为 | |x+3| - 2| = 0
|x+3| - 2 = 0
|x+3| = 2
∴ x + 3 = ±2
得 x = -1 或 x = -5
(iii) k > 0, 则方程为 | |x+3| - 2| = k
|x+3| - 2 = ±k
|x+3| = k + 2 (k + 2必>0) 或 |x+3| = -k + 2 (令- k + 2 ≧ 0, 才有解)
x + 3 = ±(k + 2) x + 3 = ±(-k + 2)
x = k - 1 或 x = -k - 5 x = -k - 1 或 x = = k - 5 (k ≦ 2)
总结(i), (ii), (iii),得
1. 当 k < 0, 方程无解
2. 当 k = 0, x = -1 或 x = -5
3. 当 0 < k < 2, x = k - 1 或 x = -k - 5 或 x = -k - 1 或 x = = k - 5
4. 当 k = 2, x = -7 或 x = -3 或 x = 1
5. 当 k > 2, x = k - 1 或 x = -k - 5