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[函数] 根式化简

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hejoseph Posted 2023-1-16 14:20 |Read mode
Last edited by hejoseph 2023-1-16 15:01已知 $0\leqslant a\leqslant b$,$0\leqslant a\leqslant c$,$2(ab+ac+bc)\geqslant a^2+b^2+c^2$,化简
\begin{align*}
&\sqrt{-2b^2+b(c+a)-(c-a)^2+2b\sqrt{a^2+b^2+c^2-ab-ac-bc}}\\
&{}+\sqrt{-2c^2+c(a+b)-(a-b)^2+2c\sqrt{a^2+b^2+c^2-ab-ac-bc}}
\end{align*}

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kuing Posted 2023-1-16 15:44
当 `a` 不是最小的时候似乎也能化简,结果有所不同,是
\[\sqrt{-2a^2-5b^2-5c^2+5a(b+c)+2bc+(4b+4c-2a)\sqrt{a^2+b^2+c^2-ab-ac-bc}}.\]

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 Author| hejoseph Posted 2023-1-16 15:59
kuing 发表于 2023-1-16 15:44
当 `a` 不是最小的时候似乎也能化简,结果有所不同,是
\[\sqrt{-2a^2-5b^2-5c^2+5a(b+c)+2bc+(4b+4c-2a)\s ...
是的,只是结论不那么好看

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 Author| hejoseph Posted 2023-2-20 14:12
无人做,直接发结论,化简结果是
\[
\sqrt{-2a^2+a(b+c)-(b-c)^2+2a\sqrt{a^2+b^2+c^2-ab-ac-bc}}
\]

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hbghlyj Posted 2023-9-15 18:17
如何化简呢

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