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  #16  
Old 19 ¾ÄȨԡÒ¹ 2009, 07:46
banker banker äÁèÍÂÙèã¹Ãкº
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Çѹ·ÕèÊÁѤÃÊÁÒªÔ¡: 24 Á¡ÃÒ¤Á 2002
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áÅéÇẺ¹ÕéÅèÐ ¼Ô´µÃ§ä˹

$-1 = -1$

$\frac{-1}{1} = \frac{1}{-1} $

$\sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1} } $

$\frac{\sqrt{-1} }{\sqrt{1} } = \frac{\sqrt{1}} {\sqrt{-1}} $

$(\sqrt{1}\sqrt{-1})\frac{\sqrt{-1} }{\sqrt{1} } = (\sqrt{1}\sqrt{-1}) \frac{\sqrt{1}} {\sqrt{-1}} $

$\sqrt{-1} \sqrt{-1} = \sqrt{1} \sqrt{1} $

$(i)(i) = (1)(1)$

$i^2 = 1$

$-1 = 1$
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µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #17  
Old 19 ¾ÄȨԡÒ¹ 2009, 08:10
¤usÑ¡¤³Ôm's Avatar
¤usÑ¡¤³Ôm ¤usÑ¡¤³Ôm äÁèÍÂÙèã¹Ãкº
à·¾ÂØ·¸ì
 
Çѹ·ÕèÊÁѤÃÊÁÒªÔ¡: 28 ÁÕ¹Ò¤Á 2008
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¤usÑ¡¤³Ôm is on a distinguished road
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ÍéÒ§ÍÔ§:
¢éͤÇÒÁà´ÔÁà¢Õ¹â´Â¤Ø³ banker View Post
áÅéÇẺ¹ÕéÅèÐ ¼Ô´µÃ§ä˹

$-1 = -1$

$\frac{-1}{1} = \frac{1}{-1} $

$\sqrt{\frac{-1}{1}} = \sqrt{\frac{1}{-1} } $

$\frac{\sqrt{-1} }{\sqrt{1} } = \frac{\sqrt{1}} {\sqrt{-1}} $

$(\sqrt{1}\sqrt{-1})\frac{\sqrt{-1} }{\sqrt{1} } = (\sqrt{1}\sqrt{-1}) \frac{\sqrt{1}} {\sqrt{-1}} $

$\sqrt{-1} \sqrt{-1} = \sqrt{1} \sqrt{1} $

$(i)(i) = (1)(1)$

$i^2 = 1$

$-1 = 1$
ºÃ÷ѴÊÕÊéÁ à¾ÃÒÐ à»ç¹ $ i=-i$
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µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #18  
Old 21 ¾ÄȨԡÒ¹ 2009, 16:35
~king duk kong~'s Avatar
~king duk kong~ ~king duk kong~ äÁèÍÂÙèã¹Ãкº
¡ÃкÕè»ÃÐÊÒ¹ã¨
 
Çѹ·ÕèÊÁѤÃÊÁÒªÔ¡: 26 ¡Ã¡®Ò¤Á 2009
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~king duk kong~ is on a distinguished road
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áÅéÇÍѹ¹ÕéÅèФÃѺ
$x^4+x^3+x^2+x+1=0$
$x(x^3+x^2+x+1)+1=0$
$x^3+x^2+x+1=\frac{-1}{x}$
$x^4+\frac{-1}{x}=0$
$\frac{x^5-1}{x}=0$
$x=1$
¼Ô´µÃ§ä˹ËÇèÒ
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  #19  
Old 21 ¾ÄȨԡÒ¹ 2009, 16:41
Scylla_Shadow's Avatar
Scylla_Shadow Scylla_Shadow äÁèÍÂÙèã¹Ãкº
ÅÁ»ÃÒ³äÃéÊÀÒ¾
 
Çѹ·ÕèÊÁѤÃÊÁÒªÔ¡: 10 ¡ØÁÀҾѹ¸ì 2009
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Scylla_Shadow is on a distinguished road
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ÍéÒ§ÍÔ§:
¢éͤÇÒÁà´ÔÁà¢Õ¹â´Â¤Ø³ ~king duk kong~ View Post
áÅéÇÍѹ¹ÕéÅèФÃѺ
$x^4+x^3+x^2+x+1=0$
$x(x^3+x^2+x+1)+1=0$
$x^3+x^2+x+1=\frac{-1}{x}$
$x^4+\frac{-1}{x}=0$
$\frac{x^5-1}{x}=0$
$x=1$
¼Ô´µÃ§ä˹ËÇèÒ
àÃÔèÁ¼Ô´µÃ§ºÃ÷ѴÊÕá´§¤ÃѺ
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
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