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[不等式] 比较大小

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人间风雪客 posted 2023-11-27 21:37 |Read mode
各位看看这道题,比较X,Y,Z的大小,能否帮忙写下过程呢?感谢感谢
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realnumber posted 2023-11-28 15:20
\[2\sqrt{log_{19}18\times log_{19}20}\le log_{19}18+log_{19}20=log_{19}360<2 \]
所以$log_{19}18\times log_{19}20< 1$即x>y
z>$\frac{19}{18}>x$
即要证\[18^{\frac{1}{18}}>19^{\frac{1}{19}}\]
即\[\frac{1}{18}\ln 18>\frac{1}{19}\ln 19\]
构造函数$y=\frac{\ln x}{x}$,求导,在x>e单调递减

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多谢大佬  posted 2023-12-3 11:52

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