若方程x3-x+1=0在区间（a，b）上有一根，其中a，b是整数，且b-a=1，则a+b=______．
令f（x）=x3-x+1，
方程x3-x+1=0在区间（a，b）上有一根，即函数f（x）在区间（a，b）内有一零点，
把x=-2，0，1，2，代入验证，
由零点存在定理知，若f（a）•f（b）＜0，则在（a，b）内存在零点，
计算知f（-2）＜0，f（-1）＞0，
所以零点在（-2，-1）内，又b-a=1，
∴a=-2，b=-1，
则a+b=-3，
故答案为：-3．
| 看图选择最佳答案。 |
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