题目内容
已知f（x+
）=x3+
，则f（x）=
| 1 |
| x |
| 1 |
| x3 |
x3-3x
x3-3x
．分析：由f（x+
）=x3+
=(x+
)(x2 -1+
)=(x+
)[(x+
)2- 3]，能求出f（x）．
| 1 |
| x |
| 1 |
| x3 |
| 1 |
| x |
| 1 |
| x2 |
| 1 |
| x |
| 1 |
| x |
解答：解：∵f（x+
）=x3+
=(x+
)(x2 -1+
)
=(x+
)[(x+
)2- 3]
=(x+
)3-3(x+
)，
∴f（x）=x3-3x．
故答案为：x3-3x．
| 1 |
| x |
| 1 |
| x3 |
=(x+
| 1 |
| x |
| 1 |
| x2 |
=(x+
| 1 |
| x |
| 1 |
| x |
=(x+
| 1 |
| x |
| 1 |
| x |
∴f（x）=x3-3x．
故答案为：x3-3x．
点评：本题考查函数解析式的求解及常用方法，是基础题．解题时要认真审题，仔细解答．
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