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  #1  
Old 01 ¡Ã¡®Ò¤Á 2012, 18:10
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͹ءÃÁ͹ѹµì˹Öè§à»ç¹Í¹Ø¡ÃÁàâҤ³Ôµ ·ÕèÁռźǡ¢Í§Í¹Ø¡ÃÁà·èҡѺ3 ¶éÒa,b,c à»ç¹ÊÒÁ¾¨¹ìáá¢Í§Í¹Ø¡ÃÁ¢éÒ§µé¹
áÅÐ 2a,3b,4cà»ç¹ÅӴѺàÅ¢¤³Ôµ áÅéǼŵèÒ§ÃèÇÁ¢Í§ÅӴѺàÅ¢¤³Ôµ¹Õéà·èҡѺà·èÒã´(ÁÍ. 53)

1. $\frac{-3}{4}$

2. $\frac{-1}{4}$

3. $\frac{1}{4}$

4. $\frac{3}{4}$

Åͧ¤Ô´áÅéǤӵͺÁѹäÁèµÃ§ÍèФèÐ ¤Ô´¼Ô´á¹èæàÅ ú¡Ç¹ªèÇ·չФР¢Íº¤Ø³ÁÒ¡¤èÐ
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  #2  
Old 01 ¡Ã¡®Ò¤Á 2012, 18:46
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ÍèÒ ¢Íº¤Ø³ÁÒ¡æàŹФÐ

áËÐæàÃÒËÒÁÒá¤è¤èÒ r ¹Ñè¹àͧ ¹Ö¡ÇèҤӵͺ

01 ¡Ã¡®Ò¤Á 2012 18:57 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ BuMMxz
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
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Old 01 ¡Ã¡®Ò¤Á 2012, 18:48
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$\frac{a}{1-r} = 3 $ â´Â $\left|r\,\right| <1 ...........(1)$
$a,b,c$ à»ç¹ÊÒÁ¾¨¹ìáá
$b = ar$ ,$ c= ar^2$
$2a , 3b ,4c$ à»é¹ÅӴѺàÅ¢¤³Ôµ
$\frac{2a+4c}{2} = 3b$
$a+2c = 3b$
$a+2ar^2 = 3ar$
$1+2r^2 = 3r$
$(2r-1)(r-1) = 0$
$r= \frac{1}{2} $, $1$ áµè $ \left|r\,\right| <1$
$\therefore r=\frac{1}{2}$
¨Ò¡$ (1)$ ¨Ðä´é $a = \frac{3}{2}$ ,$ b = \frac{3}{4}$
¼ÅµèÒ§ÃèÇÁ $= 3b-2a = \frac{9}{4} - 3 =\frac{-3}{4}$

01 ¡Ã¡®Ò¤Á 2012 18:48 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Euler-Fermat
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #4  
Old 01 ¡Ã¡®Ò¤Á 2012, 19:04
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