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àŢ¡¡ÓÅѧ
1. ¨§ËÒ¤èÒ X ¨Ò¡ÊÁ¡Òà $4^ 2x+1 $ = $32^ x-3 $
2. (0.125)(0.5)^2(0.25)^3 à·èҡѺà·èÒäËÃè 3. 49$^\frac{1}{2} $ + 32$^\frac{4}{5}$ ÁÕ¤èÒà·èҡѺà·èÒäËÃè 4. $\frac{2^n+1}{(2^n)^n-1}$ $\div$ $\frac{4^n+1}{(2^n)^n}$ ÁÕ¤èÒà·èҡѺà·èÒäËÃè
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1) ⨷Âìà»ç¹áººä˹¤ÃѺ
$4^{2x+1}=32^{x-3}$ ¶éÒ $4^{2x+1}=32^{x-3}$ $2^{2(2x+1)}=2^{5(x-3)}$ $2(2x+1)=5(x-3)$ $x=17$ 2) $(\dfrac{125}{1000})(\dfrac{25}{100})(\dfrac{25}{100})(\dfrac{25}{100})(\dfrac{25}{100})=(\dfrac{1}{8})(\dfrac{1}{4})(\dfrac{1}{ 64})=\dfrac{1}{2048}$ 3) $7+2^4=7+16=23$ 4)
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17 ÊÔ§ËÒ¤Á 2009 11:21 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 4 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Ne[S]zA |
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3. à»ÅÕè¹ 49 áÅÐ 32 ãËéà»ç¹àŢ¡¡ÓÅѧáÅéǨеѴ仴ͧ
2.à»ÅÕè¹°Ò¹à»ç¹ 0.5 4. $2^{n+1}=2^n*2^1$
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1 = 2 ä´é 555+ ÁѹäÁèÁÕÍÐäÃá¹è¹Í¹ 555+ |
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äÁèà¢éÒ㨢éÍ2.¤Ð
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à¾ÃÒÐÇèÒ $(0.125)\times(0.5)^2\times(0.25)^3$
= $(125\times10^{-3})\times(5\times10^{-1})^2\times(25\times10^{-2})^3$ = $(5^3\times10^{-3})\times(5\times10^{-1})^2\times(5^2\times10^{-2})^3$ = $(5^3\times5^2\times5^6)(10^{-3}\times10^{-2}\times10^{-6})$ = $(5^{11})(10^{-11})$ = $(5^{11})(5^{-11}\times2^{-11})$ = $2^{-11}$ = $\frac{1}{2048}$ »Å. ¡ÒÃà·¤ log ¢Í§¤Ø³ à¹Ê ´Ùá»Å¡æ¹Ð¤ÃѺ 16 ÊÔ§ËÒ¤Á 2009 17:01 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ [SIL] à˵ؼÅ: à¾ÔèÁ »Å. |
#6
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ÍëÍ à¢éÒã¨áÅéǤÐ
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¢éÍ1. ¶éèÒ⨷Âìà»ç¹
¨§á¡éÊÁ¡Òà $4^{2x}+1=32^x-3$ ¨Ðä´é¤ÓµÍº´Ñ§ÀҾṺ (¨Ò¡ wolfram alpha) |
#8
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§èÐ ¨ÃÔ§´éÇ ·Óä»ä´éä§ËÇèÒ ¡ÃШÒÂä»àÅ àËÍææ
¢Íº¤Ø³·Õèàµ×͹¤ÃѺ (ź´èǹÍÔÍÔ)
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#9
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ÁÒªèÇÂà´Ò¤ÃѺ ¤Ô´ÇèÒ⨷Â줧à»ç¹ $4^{2x+1}=32^{x-3}$
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ÍéÒ§ÍÔ§:
$=\dfrac{2 \cdot 2^n \cdot 2^n}{(2^n)^n} \times \dfrac{(2^n)^n}{2^{2n} \cdot 4}$ $=\dfrac{2 \cdot 2^{2n}} {(2^n)^n} \times \dfrac{(2^n)^n}{2^{2n} \cdot 4}$ $=\dfrac{1}{2}$
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) |
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