#1
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ÊÍǹ. ¤èÒ 2
¾ÔÊÙ¨¹ì àÍ¡Åѡɳì
(A^3+B^3+C^3)-3ABC=(A+B+C)(A^2+B^2+C^2-AB-BC-CA) â´ÂãªéÇÔ¸Õ·Ò§ àâҤ³Ôµ |
#2
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- -* ¹Õè⨷ÂìÃдѺªÑé¹ÍÐäÃËÃͤÃѺ
§§ÁÒ¡æàÅÂÍÐ T T |
#3
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¡ç$(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ ¡çÃÙéæ¡Ñ¹¹Ð¤ÃѺ áµèäÍé¾ÔÊÙ¨¹ì·Ò§àâÒà¹ÕèÂÂÒ¡¨ÃÔ§æ ¶éÒà»ç¹»Ô·Ò¡ÍÃѸ¡çÇèÒä»ÍÂèÒ§ ÎèÒæ
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I think you're better than you think you are. |
#4
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·ÒÁÁÑÂ⨷Âì¶Ö§ÂÒ¡¨Ñ§¤èÐ
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#5
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µÑé§ã¨µèÍ仹ÐÊÙéæ
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#6
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.ã¤Ã¾ÔÊÙ¨¹ìà»ç¹ºéÒ§¤Ñº
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#7
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áËÁè... äÍé´Õ¡ÃÕÊͧ¹ÕèÂѧ¾Í¹Ö¡ÃÙ»ä´é
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#8
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àÍ¡Åѡɳì¡ÓÅѧÊÒÁà¹Õè ãªéÅÙ¡ºÒÈ¡ìªèÇÂä´é»Ð¤ÃѺ ???
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#9
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¢Í»Åء˹èͤÃѺ ÂѧäÁèÁÕã¤Ãà©ÅÂàÅÂ
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#10
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¾ÔÊÙ¨¹ì¤ÃѺ
¨Ò¡ $(A+B)^3 = A^3+3A^2B+3AB^2+B^3$ $A^3+B^3+C^3 = (A+B)^3-3AB(A+B)+c^3-3ABC$ $=[(A+B)^3+C^3]-3AB(A+B)-3ABC$ $=(A+B+C)[(A+B)^2-(A+B)C+C^]-3AB(A+B+C)$ $=(A+B+C)(A^2-AB+B^2-AC-BC+C^2)$ ¶éÒãªéàÃҢҨзÓ䧤ÃѺ ¼ÁÇèÒ¾ÔÊÙ¨¹ì·Ò§¾Õª¤³Ôµ´Ù§èÒ¡ÇèÒÍÕ¡ 04 Á¡ÃÒ¤Á 2009 19:42 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 4 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¤³ÔµÈÒʵÃì |
#11
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¼ÁÇèÒ¹èÒ¨ÐàÅ蹡ѺÅÙ¡ºÒÈ¡ì·ÕèÁÕ¤ÇÒÁÂÒÇ´éÒ¹ $A+B+C$ ˹èǤÃѺ
áÅéÇ¡çãªéà·¤¹Ô¤áºº double counting ã¹ combinatorics ¤×ÍÁͧ¢Í§ÊÔè§à´ÕÂǡѹáµèÁͧÊͧẺ ¼ÁÂѧäÁèä´éÅͧ¤Ô´ á¤èà´ÒàÍÒ¤ÃѺ
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site:mathcenter.net ¤Ó¤é¹ |
#12
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à¾ÔèÁàµÔÁ¹Ô´¹Ö§¹Ð¤ÃѺ
¶éÒ $A+B+C = 0$ ¨Ðä´é $A^3+B^3+C^3 = 3AB$ ¤ÃѺ
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#13
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µ¡ä»µÑǹ֧ËÃ×Í»ÅèÒ¤ÃѺ
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