#1
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àŢ¡¡ÓÅѧ
[2(9^x-1)]+[2^x-1/2] = [2^x+1/2]+[3^2x-1]
x=? |
#2
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$[2(9^{x-1})]+[2^{x-0.5}] = [2^x+1/2]+[3^{2x-1}]$ ËÃ×Í
$[2(9^{x}-1)]+[2^{x}-1/2] = [2^{x}+1/2]+[3^{2x}-1]$¤ÃѺº º º ?
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13 ÁԶعÒ¹ 2012 21:46 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¤usÑ¡¤³Ôm |
#3
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µÑǺ¹¤ÃѺ ^^
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#4
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ضéÒäÁèä´éáÊ´§ÇèÒ¼ÁÅ͡⨷Âì¼Ô´ = =
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#5
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⨷Âìà»ç¹áºº¹ÕéËÃ×Íà»ÅèÒ¤ÃѺ
$2(9^{x-1}) +2^{x-\frac{1}{2}} = 2^{x+\frac{1}{2}}+3^{2x-1}$ 13 ÁԶعÒ¹ 2012 22:07 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Euler-Fermat |
#6
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*2^x-1/2 ¤ÃѺ
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#7
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á¡éáÅéǤÃѺ áµèãªèẺ¹ÕéËÃ×Íà»ÅèÒ¤ÃѺ
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#8
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¤ÃѺ ^^ ¹èÒ¨ÐäªèáÅÐ
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#9
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$2(9^{x-1})+2^{x-\frac{1}{2}} = 2^{x+\frac{1}{2}} +3^{2x-1}$
$\frac{2}{9}(9^x) +\frac{1}{\sqrt{2}}(2^x) = \sqrt{2}(2^x) +\frac{1}{3}(9^x)$ $\frac{1}{\sqrt{2}}(2^x) - \sqrt{2}(2^x) = \frac{1}{9}(9^x)$ $\frac{-1}{\sqrt{2}}(2^x) = \frac{1}{9}(9^x)$ $\frac{-9}{\sqrt{2}} = (\frac{9}{2})^x$ «Öè§$ (\frac{9}{2})^x >0 $ äÁèÁÕ¤èÒ x ·Õà»ç¹¨Ó¹Ç¹¨ÃÔ§·ÕèÊÍ´¤Åéͧ 13 ÁԶعÒ¹ 2012 22:15 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Euler-Fermat |
#10
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¢Íº¤Ø³¤ÃѺ ^^
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#11
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¤Ë9.ÇÔ¸Õ¹èҨжء¤ÃѺ(äÁèá¹èã¨àËÁ×͹¡Ñ¹)
áµè¤ÓµÍº¼ÁÇèҵͺ ૵ÇèÒ§ ¹Ð ^^ 13 ÁԶعÒ¹ 2012 22:30 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Teh |
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