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#1
15 ÊÔ§ËÒ¤Á 2010, 19:26
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ÊÁ¡ÒáÓÅѧ3
¼ÁÍÂÒ¡ËÒÃÒ¡àªè¹(ÊÁ¡ÒáÓÅѧ2)
$ax^2+bx+c $ ãËé y,z à»ç¹ÃÒ¡¢Í§ÊÁ¡Òà $y+z=-\frac{b}{a}$ $y\times z=\frac{c}{a} $ ÊÁ¡ÒáÓÅѧ 3 ÁÕÊÙµÃäËÁ¤ÃѺ  15 ÊÔ§ËÒ¤Á 2010 19:26 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¤¹ÍÂÒ¡à¡è§ 
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#2
15 ÊÔ§ËÒ¤Á 2010, 20:10
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ÍéÒ§ÍÔ§:
$ax^3+bx^2+cx+d=0$ $p+q+r=-b/a$ $pq+qr+rs=c/a$ $pqr = -d/a$
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#3
15 ÊÔ§ËÒ¤Á 2010, 20:19
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áÅéǨоÔÊÙ¨¹ìÍÂèÒ§äÃËÃͤÃѺ
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** ¶éÒäÁèÊÙé¨ÐÃÙéËÃ×ÍÇèÒá¾é ¶éÒÍè͹áͤ§äÁèÃÙéÇèÒà¢éÁá¢ç§ ** äÁèÂ×¹ËÂÑ´¤§äÁèÃÙéÇèÒÁÕáç äÁè¶Ù¡á«§¤§äÁèÃÙéàÃÒªéÒä» ** Sub #1 ÊÔ觷ÕèÁÑè¹ã¨·ÕèÊØ´¡ÅѺ·ÓãËéÃÙéÊÖ¡áÂè·ÕèÊØ´ T T
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#4
15 ÊÔ§ËÒ¤Á 2010, 20:35
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ÍéÒ§ÍÔ§:
¨Ðä´é $ (x-p)(x-q)(x-r) =0 $ $x^3-(p+q+r)x^2+(pq+qr+pr)x-pqr =0$ = $ ax^3+bx^2+cx+d $ ËÒÃÊÁ¡ÒâÇÒ´éÇ a µÅÍ´ ; $x^3-(p+q+r)x^2+(pq+qr+pr)x-pqr =0$ = $ x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} $ à·Õº ÊÑÁ»ÃÐÊÔ·¸Ôìä´éÇèÒ $p+q+r = -\frac{b}{a}$ $pq+qr+pr = \frac{c}{a}$ $pqr = -\frac{d}{a}$ 15 ÊÔ§ËÒ¤Á 2010 20:36 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ -Math-Sci- à˵ؼÅ: Å×Á x *
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#5
15 ÊÔ§ËÒ¤Á 2010, 21:56
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¶ÒÁÍÕ¡ÍÂèÒ§¤ÃѺ
ÃÒ¡¢Í§ÊÁ¡Òà ¡Ñº ¤ÓµÍº¢Í§ÊÁ¡Òà µèÒ§¡Ñ¹äËÁ¤ÃѺ ¢Íº¤Ø³¤ÃѺ
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#6
15 ÊÔ§ËÒ¤Á 2010, 22:06
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äÁèµèÒ§¤ÃѺ
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¤³ÔµÈÒʵÃì ¤×Í ÀÒÉÒÊÒ¡Å ¤³ÔµÈÒʵÃì ¤×Í ¤ÇÒÁÊǧÒÁ ¤³ÔµÈÒʵÃì ¤×Í ¤ÇÒÁ¨ÃÔ§ µÔ´µÒÁªÁ¤ÅÔ»ÇÕ´ÕâÍä´é·Õèhttp://www.youtube.com/user/poperKM
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#7
15 ÊÔ§ËÒ¤Á 2010, 22:21
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¢Íº¤Ø³ÁÒ¡¤ÃѺÊÓËÃѺ·Ø¡¤ÓµÍº.
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