悠闲数学娱乐论坛(第3版)»论坛 › 数学区 › 初等数学讨论 › 曲线的二重切线数$\frac{n}{2}(n-2)\left(n^2-9\right)$

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[几何] 曲线的二重切线数$\frac{n}{2}(n-2)\left(n^2-9\right)$

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hbghlyj Posted at 2025-2-17 04:58:29 |Read mode

$n$次曲线的最多的二重切线数$=\frac{n}{2}(n-2)\left(n^2-9\right)$，如何证明

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Author| hbghlyj Posted at 2025-2-19 00:19:45

何时取等

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