weilingyou
幼苗
共回答了17个问题采纳率:70.6% 举报
(1)楼上
(2)因为f(1)=2得 a + 1/b + c = 2,
而且a,b,c∈Z,
所以b = 1, a + c = 1
=> a = 0, b = 1, c = 1 或者 a = 1, b = 1, c = 0
由f(2) < 3得 4a + 1/2b + c < 3
将a = 0, b = 1, c = 1代入上式,成立
将a = 1, b = 1, c = 0代入上式,显然不成立
综上:a = 0, b = 1, c = 1
1年前
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