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  #1  
Old 10 ¸Ñ¹ÇÒ¤Á 2009, 19:29
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$3^x=4^y=5^z=60^k$áÅéÇ$\frac{1}{y} +\frac{1}{x}+\frac{1}{z} $ ÁÕ¤èÒà·èÒäÃ
(math contest Benchama 2008)
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  #2  
Old 10 ¸Ñ¹ÇÒ¤Á 2009, 19:49
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ä´é $\frac{1}{k}$ ¤ÃѺ
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  #3  
Old 10 ¸Ñ¹ÇÒ¤Á 2009, 20:27
ÍÂÒ¡à¡è§àÅ¢¤ÃѺ's Avatar
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$3=60^\frac{k}{x}$
$4=60^\frac{k}{y}$
$5=60^\frac{k}{z}$

$3x4x5=60^{k(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}$
$60^1=60^{k(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}$
$1=k(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{k}$
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #4  
Old 10 ¸Ñ¹ÇÒ¤Á 2009, 20:31
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¢Íº¤Ø³·Ø¡¤¹ÁÒ¡¹Ð¤ÃѺ ^_^
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Old 10 ¸Ñ¹ÇÒ¤Á 2009, 23:35
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ÊͺÁÐÃ×¹¹ÕéáÅéÇ à¤Ã×èͧàÂç¹à¨ÕêºàŤÃѺ
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  #6  
Old 11 ¸Ñ¹ÇÒ¤Á 2009, 00:04
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ÍéÒ§ÍÔ§:
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ÊͺÁÐÃ×¹¹ÕéáÅéÇ à¤Ã×èͧàÂç¹à¨ÕêºàŤÃѺ
¢ÍãËé⪤´Õ¹Ð¤ÃѺ (¾Õè¼Á¡çàªè¹¡Ñ¹¤ÃѺ)
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