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[函数] 函数题有没有问题

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依然饭特稀 Posted at 2013-10-18 18:12:04 |Read mode
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kuing Posted at 2013-10-18 18:36:02
还是没看懂题,是对 i=1 和 i=2 的三角形都要成立还是只要对两者之一成立就可以?

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其妙 Posted at 2013-10-18 18:54:05
回复 2# kuing
总觉得题意有点不明朗,
妙不可言,不明其妙,不着一字,各释其妙!

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 Author| 依然饭特稀 Posted at 2013-10-20 08:17:53
我的疑惑也就是这,显然不可能i=1和i=2同时成立,若理解为其一成立,当i=1时满足f(0)=f(2)≠f(1),f(3)任意共4*3*4=48,i=2时是6*4=24,总共是72.

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