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#1
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#2
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¨Ò¡ $60^{1-b}=12$ ¨Ðä´é $12^{1/(2(1-b))}=60^{1/2}$
´Ñ§¹Ñé¹ $12^{(1-a-b)/(2(1-b))}=(60^{1-a-b})^{1/2}=\sqrt{4}=2$ »Å. ¼Á¨Óä´éÇèÒà¤ÂµÍº¤Ó¶ÒÁ¢é͹Õéàͧ㹡ÃзÙé¢éÍÊͺä˹äÁèÃÙé ¨ÓäÁèä´é ⪤´Õ·Õè¢é͹ÕéÇÔ¸Õ·ÓÊÑé¹ Âѧ䧤ÃÒÇ˹éÒ¡è͹µÑ駡ÃзÙéãËÁèú¡Ç¹¤é¹¡ÃзÙéÊÑ¡¹Ô´ ËÃ×ÍäÁè¡çºÍ¡Êѡ˹èÍÂÇèÒä´é⨷ÂìÁÒ¨Ò¡ä˹¹Ð¤ÃѺ
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¤¹ä·ÂÃèÇÁã¨ÍÂèÒãªéÀÒÉÒÇÔºÑµÔ ½Ö¡¾ÔÁ¾ìÊÑÅѡɳìÊÑ¡¹Ô´ ªÕÇÔµ(¤¹µÍºáÅФ¹¶ÒÁ)¨Ð§èÒ¢Öé¹àÂÍÐ (¨ÃÔ§æ¹Ð) Stay Hungry. Stay Foolish. 20 ¡ØÁÀҾѹ¸ì 2008 21:55 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ nongtum |
#3
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ÍÂÒ¡·ÃÒºÇèÒ ºÃ÷Ѵáá·ÕèºÍ¡ÇèÒ
601-b = 12 à¹Õè ÁÒ¨Ò¡ä˹ËÃͤÐ?? 21 ¡ØÁÀҾѹ¸ì 2008 09:04 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Na-na*Zz |
#4
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$60^{1-b}=60/60^b=60/5=12$ ¤ÃѺ ËÒ¡·´µÒÁ¨ÐàËç¹ÇèÒÁÕÍÕ¡¨Ø´·Õèãªé¡Ãкǹ¡ÒÃà´ÕÂǡѹ¹Õé
ÍÂèÒÅ×ÁÇèÒ¢é͹ÕéÁÕÊèǹ·Õè¡¡ÓÅѧà»ç¹àÈÉÊèǹ´éǹФÃѺ
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¤¹ä·ÂÃèÇÁã¨ÍÂèÒãªéÀÒÉÒÇÔºÑµÔ ½Ö¡¾ÔÁ¾ìÊÑÅѡɳìÊÑ¡¹Ô´ ªÕÇÔµ(¤¹µÍºáÅФ¹¶ÒÁ)¨Ð§èÒ¢Öé¹àÂÍÐ (¨ÃÔ§æ¹Ð) Stay Hungry. Stay Foolish. |
#5
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¢ÍªèÇÂá¡é¢éÍʧÊÑ´éǹФÃѺ
⨷Âìµéͧ¡ÒÃËÒ¤èÒ $12^\frac{(1−a−b)}{2(1−b)}$ â´ÂãËé¤èҢͧ$60^a$ áÅÐ$60^b$ ÁÒ áÊ´§ÇèÒàÃÒ¤ÇÃá»Å§â¨·ÂìãËéÍÂÙèã¹ÃÙ» $60^n$ ¨Ö§¨Ð¤Ô´ä´é§èÒ¢Öé¹ à¹×èͧ¨Ò¡ 12 = $\frac{60}{5}$ = $\frac{60}{60^b}$ = $60^{(1-b)}$ àËÁ×͹·Õè¤Ø³Nongtum áÊ´§äÇéáÅéǹÑè¹àͧ áÅéǨѺä»á·¹¤èÒàÅ¢12 ã¹â¨·Âì ¨Ðä´é $12^\frac{(1−a−b)}{2(1−b)}$ = $(60^{(1-b)})^\frac{(1−a−b)}{2(1−b)}$ = $60^{(1-b).\frac{(1−a−b)}{2(1−b)}}$ = $60^{\frac{(1−a−b)}{2}}$ = $(\frac{60^1}{60^a.60^b})^{\frac{1}{2}}$ = $(\frac{60}{3x5})^{\frac{1}{2}}$ áÅéÇ¡ç·Óµè͵ÒÁẺ¤Ø³ Nongtum ¡ç¨Ðä´é¤ÓµÍº¤ÃѺ 27 ¡ØÁÀҾѹ¸ì 2008 00:54 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Puriwatt à˵ؼÅ: ¨Ñ´ãËé´Ù§èÒ¢Ö鹤ÃѺ |
#6
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#7
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555+ ¢é͹Õéä˧¼ÁÍÑ¡ Log «Ð§Ñé¹àÅ à«èÍÊØ´ææ ¤Ô´ä´é䧿ÃÐ ÍÑ´ Log àª×èÍàÃÒàÅéÂ
** ¢é͹Õéà»ç¹â¨·Âì ÊÊÇ· 2550 Ãͺáá¤ÃѺ ¢é͹Õéà»ç¹â¨·Âì IMC ÃдѺࢵ»Õ 2550 ´éǤÃѺ |
#8
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