#1
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àŢ¡¡ÓÅѧ
¼ÁʧÊÑÂÍèÒ¤ÃѺ ÇèÒ (-1)^(6/2) ÁѹÁÕ¤èÒà·èÒäËÃèÍФÃѺ
¤×ͺҧ¤¹à¤éҺ͡ÇèÒ áºº·Õè 1 : (-1)^(6/2) = (-1)^3 = -1 ºÒ§¤¹¡çºÍ¡ÇèÒ áºº·Õè 2 : (-1)^(6/2) = [(-1)^6]^(1/2) = 1^(1/2) = 1 ÊÃØ»áÅéÇÁѹà·èҡѺ 1 ËÃ×Í -1 ¡Ñ¹á¹èÍФÃѺ ¢Íâ·É´éǤÃѺ¾ÔÁ latex äÁè໹ÍèÒ¤ÃѺ |
#2
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Ẻ·Õè 1 ¶Ù¡¤ÃѺ
$1^{\frac{1}{2}}$ à»ç¹ä´é·Ñé§ $1,-1$ ¹Ð¤ÃѺ
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16.7356 S 0 E 18:17:48 14/07/15 04 àÁÉÒ¹ 2014 17:52 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Sirius |
#3
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#4
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¨Ò¡·Õèà¢Õ¹ÁÒÁÕǧàÅçºà¹é¹·ÕèàŢ¡¡ÓÅѧÇèÒà»ç¹$(\frac {6}{2}) $
¹èҨзÓã¹Ç§àÅ纡è͹à»ç¹ $(-1)^3=-1$ áµè¶éÒà»ç¹$(-1)^{\frac {6}{2}} $ ¨Ò¡¡®¢éÍ·Õè9¢Í§àŢ¡¡ÓÅѧ $a^{\frac{m}{n}}=\sqrt[n]{a^{m}} $ ´Ñ§¹Ñ鹨ҡ⨷Âì¨Ðä´é $=\sqrt{(-1)^{6}} =1$ |
#5
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ÁÕà§×è͹ä¢ÍÐäúéÒ§äËÁ¤ÃѺ
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#6
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à§×è͹䢤×Í $a\not= 0\,$áÅÐ$\,m\in \mathbf{N} ;n\geqslant 2$
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