#1
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àŢ¡¡ÓÅѧ
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1.$3(30)^x-6(15)^x-3(6)^x+2(5)^x-10^x+2^x-2=0$ 2.$16^x+36^x=2\times 81^x$ 3.$9^{x^2-1}-(36\times 3^{x^2-3})+3=0$ ¢éÍ3 ¼Á¤Ô´ä´é¶Ö§ $3^{x^2} = \frac{1}{10}$ áÅéÇ仵èÍäÁèà»ç¹ÍÐ - -"
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#2
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ÍéÒ§ÍÔ§:
$9^{x^2-1}-(36\times 3^{x^2-3})+3=0$ $ \frac{3^{2x^2}}{9} - \frac{36 \cdot 3^{x^2}}{3^3} +3 =0$ $3^{2x^2} - 12 \cdot 3^{x^2} +27 =0$ $(3^{x^2}-3)(3^{x^2} -9) =0$ $3^{x^2} = 3^1, \ \ 3^2$ $x^2 =1, \ \ 2$ $x = \pm 1, \ \ \pm \sqrt{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´) 20 ¾ÄÉÀÒ¤Á 2012 07:59 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ banker |
#3
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¢Íº¤Ø³¤ÃѺ áÅéÇ¢éÍ 1 ¡Ñº 2 ¤Ô´Âѧ䧤ÃѺ
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#4
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$16^x+36^x=2\times 81^x$
$2^{4x} + 2^{2x} \cdot 3^{2x} - 2 \cdot 3^{4x}$ Let $ \ \ 2^{2x} = a, \ \ 3^{2x} = b$ Hence $ \ \ a^2 + ab - 2 b^2$ $ (a-b)(a+2b) = 0$ $2^{2x} = 3^{2x} \ \ \to x = 0$ x = 0 ÊÓËÃѺ¨Ó¹Ç¹¨ÃÔ§
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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´) |
#5
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¢éÍ 1 ´ÙáÅéǵÒÅÒ ¤Ô´äÁèÍÍ¡
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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´) |
#6
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ʧÊÑÂ⨷Âìà¡Ô¹ÁÒ 1 ¾¨¹ì
¶éÒá¤è¹Õé $3(30)^x-6(15)^x+2(5)^x-10^x+2^x-2=0$ $x=1$ |
#7
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ÍéÒ§ÍÔ§:
ÃÖ»èÒǤѺ - -
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#8
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¢ÍÇÔ¸Õ¤Ô´ä´é»èÒǤѺ
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#9
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#8ÅͧãËé$2^x$=a,$3^x$=b,$5^x$=cáÅéǨѴÃÙ»µèÍ¡ç¹èÒ¨Ðä´é¤ÃѺ
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#10
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ú¡Ç¹á¨§·Õ¤ÃѺ ÂѧäÁè¤èÍÂáÁè¹ÍèÒ ¨Ñ´äÁè¶Ù¡
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#11
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ãªè¤ÃѺ ÁÖ¹¨Ò¡¢éÍ 3
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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´) |
#12
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ÍéÒ§ÍÔ§:
$3(30)^x-6(15)^x+2(5)^x-10^x+2^x-2=0$ $3(2\times3\times5)^x-6(3\times5)^x+2(5)^x-(2\times5)^x+2^x-2=0$ $3(2^x\times3^x\times5^x)-6(3^x\times5^x)+2 \cdot 5^x-(2^x\times5^x)+2^x-2=0$ ãËé $ \ a = 2^x, \ \ b = 3^x, \ c = 5^x$ $3abc -6bc+2c - ac + a - 2 = 0$ $3bc(a-2) +c(2-a)+ (a-2) = 0$ $3bc(a-2) -c(a-2)+ (a-2) = 0$ $(a-2)(3bc-c+1) = 0$ $a= 2$ $2^x = 2 = 2^1$ $x = 1$ ÊèǹÍաǧàÅçº â´ÂÇÔ¸Õ Á. µé¹ ¤§·ÓäÁèä´é
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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´) |
#13
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ÍéÒ§ÍÔ§:
$3(3^x)(5^x)+1=5^x$ $3(15^x)+1=5^x$ ¶éÒ x>0 äÁèµéͧ¾Ù´ ÁÒ¡¡ÇèÒá¹è¹Í¹ ¶éÒ x=0 á·¹¤èÒáÅéÇäÁè¨ÃÔ§ ¶éÒ x<0 ¨Ðä´é LHS ÁÒ¡¡ÇèÒ 1 áµè RHS ¹éÍ¡ÇèÒ 1 à»ç¹ä»äÁèä´éàªè¹¡Ñ¹ §Ø§Ô |
#14
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¤Ô´ä´éÅФÃѺ
$3(2^x\times 3^x \times 5^x)-6(3^x \times 5^x)-3(2^x\times 3^x)+6(3^x)+2(5^x)-10^x+2^x-2=0$ ãËé $2^x=a , 3^x=b , 5^x=c$ $3ab-6bc-3ab+6b+2c-ac+a-2=0$ $3abc-3ab-6bc+6b+2c-2-ac+a=0$ $3ab(c-1)-6b(c-1)+2(c-1)-a(c-1)=0$ $(c-1)(3ab-6b-a+2)=0$ $(c-1)[3b(a-2)-1(a-2)]=0$ $(c-1)(a-2)(3b-1)=0$ $a=2 , b=\frac{1}{3} , c=1$ $x=-1,0,1$
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¨ÐÃÍ´ÁÑé¹êÍÍÍ |
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