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[函数] 求证切线斜率的倒数和为0

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lemondian posted 2024-3-8 10:25 |Read mode
这个命题如何证明:
若多项式函数$f(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n$有$n$个不同零点$x_1,x_2,\cdots x_n$,且$y=f(x)$在点$(x_i,f(x_i))$处切线的斜率为$k_i(i=1,2,\cdots ,n)$,则$\dfrac{1}{k_1}+\dfrac{1}{k_2}+\cdots \dfrac{1}{k_n}=0$。

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睡神 posted 2024-3-8 10:53
Last edited by 睡神 2024-3-8 11:12这个好像是西西大神大学时期的一个猜想,不久后他给出了证明,记忆中我保存着,但一时没找到

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看楼下,证明是k神给出的  posted 2024-3-8 12:11
除了不懂,就是装懂

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睡神 posted 2024-3-8 11:14
Last edited by 睡神 2024-3-8 12:08找到了,在此楼搬个砖
DT))5S]5Z4})(PB~MN6`5Z3.png
除了不懂,就是装懂

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kuing posted 2024-3-8 11:53
睡神 发表于 2024-3-8 11:14
找到了,在此楼搬个砖
当时他在“不等式研究 QQ 群”里发的这猜想,解答是我写给他的,见《撸题集》P.912 题目 6.8.12

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记起来了,好像你之前说过这事😂  posted 2024-3-8 11:57
将你的那个贴搬过来了,这个更原汁原味😁  posted 2024-3-8 12:09
谢谢两位!
好象还可以有更一般的结论哩  posted 2024-3-8 15:57

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