Discuz Math Forum»Forum › Mathematics › High School › $\frac1a<\frac{2π\tan \left(\frac12π\sqrt{1-8 a}\right)}{\sqrt{1-8 a}}$

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[不等式] $\frac1a<\frac{2π\tan \left(\frac12π\sqrt{1-8 a}\right)}{\sqrt{1-8 a}}$

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hbghlyj posted 2025-3-1 00:42 |Read mode

$0<a<\frac18:$\[\frac1a<\frac{2 \pi \tan \left(\frac{1}{2} \pi \sqrt{1-8 a}\right)}{\sqrt{1-8 a}}\]

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神奇的是这个8还是最大的 posted 2025-3-1 21:44

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original poster hbghlyj posted 2025-3-1 03:25

刚刚回帖遇到“不存在”的问题，非常令人困惑

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🥺 posted 2025-3-1 13:37

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2025-7-20 05:51 GMT+8

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