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¡ÒúéÒ¹ 2 ¢éͤÃѺ¼Á
1. $\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }$ ¨§·ÓãËéµÑÇÊèǹäÁèÁÕÃÒ¡
2. $\left[\,2+\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ + $\left[\,2-\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ ÁÕ¤èÒà·èҡѺà·èÒäà ¢Íº¤Ø³Åèǧ˹éÒ¤ÃѺ |
#2
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ÊÅÒ¨ҧËÒÂä»
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I think you're better than you think you are. 29 ¾ÄÉÀÒ¤Á 2008 17:49 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ RETRORIAN_MATH_PHYSICS à˵ؼÅ: äÁ觴§ÒÁ |
#3
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Hint:
¢éÍ1) ·ÓÊèǹãËéà»ç¹¼ÅºÇ¡¡ÓÅѧÊÒÁ ¢éÍ2) ÁͧáµèÅÐǧàÅçºà»ç¹µÑÇá»Ã x,y ¨Ðä´é¤ÇÒÁÊÑÁ¾Ñ¹¸ì $x^3+y^3=4=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)$ áÅéÇá¡éÊÁ¡ÒÃËÒ (x+y) <<⨷Âì¶ÒÁ x+y »Å.à´ÕëÂǹÕéÁѸÂÁ»ÅÒÂÂѧÁÕ⨷ÂìÃÙ·æÍÂÙèÍÕ¡ËÃͤÃѺà¹Õè äÁè¤èÍÂà¨Í
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I am _ _ _ _ locked 28 ¾ÄÉÀÒ¤Á 2008 18:53 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 4 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ t.B. |
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ÁÕ¤ÃѺ ÍÂÙèã¹à¹×éÍËÒ Á.5 àÃÕ¹¡è͹¿Ñ§¡ìªÑ¹àÍç¡â»¤ÃѺ
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I think you're better than you think you are. |
#5
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¤Ù³1/3 à¢éÒÁÒ¢éÒ§ã¹
=[33?231+1031?321] +[ 33?231−1031?321] ¼Á¢ÍáÂ駺Ã÷Ѵ¹Õé ¼Á¤Ô´ÇèÒ¼Ô´ ¤Ù³1/3àŢ¡¡ÓÅѧäÁä´é à¾ÃÒÐÇèÒã¹Ç§àÅçºÁÕà¤Ã×èͧËÁÒ ºÇ¡ÍÂÙè ªÇèÂàªç¤´ÙÍÕ¡¤ÃÑé§.
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#6
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¢éÍ $1$
á¹Ç¤Ô´µÒÁ·ÕèÍéÒ§ÍÔ§àŤÃѺ ¨§·ÓãËéÊèǹäÁèµÔ´ÃÒ¡ $\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }$ á¹Ç¤Ô´ ¨Ò¡ $a^{3}+b^{3}=(a+b)(a^{2}+ab+b^{2})$ áÅШҡµÑÇÊèǹ¢Í§àÃÒ ·ÕèàÃÒÁÕ¤×Í $\sqrt[3]{16}-\sqrt[3]{4}+1 $ ¨Ò¡¡ÒÃÊѧࡵØàÃÒàËç¹ÇèÒ $\sqrt[3]{16}-\sqrt[3]{4}+1=([2^{\frac{2}{3}}]^{2}-[2^{\frac{2}{3}}][1]+1^{2})$ «Öè§ÁÕÅѡɳФÅéÒÂæ $(a^{2}-ab+b^{2})$ àÃÒµéͧËÒ $(a+b)$ ÁÒàµÔÁà¢éÒ仫Ö觨зÓãËéä´éà»ç¹¼ÅºÇ¡¡ÓÅѧÊÒÁ «Ö觡ç¤×Í $(2^{\frac{2}{3}}+1)$ ¨Ò¡â¨·ÂìàÃÒ¨Ðä´é $$\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }$$ $$\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }=\frac{3}{([2^{(\frac{2}{3})}]^{2}-[2^{(\frac{2}{3})}][1]+[1^{2}])}\cdot \frac{(2^{\frac{2}{3}}+1)}{(2^{\frac{2}{3}}+1)}$$ $$\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }=\frac{3(2^{\frac{2}{3}}+1)}{(2^{\frac{2}{3}})^{3}+1^{3}}$$ $$\frac{3}{\sqrt[3]{16}-\sqrt[3]{4}+1 }=\frac{3\sqrt[3]{4}+3}{5}$$ |
#7
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ÍèÍ ¤ÃѺãªèæ ¤Ù³äÁèä´é Å×Áä»
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I think you're better than you think you are. |
#8
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¢éÍ2¼Áä´é¤ÓµÍº=2 .ãªéÇԸդԴẺ¾×èt.b. ÍÕ¡2¤ÓµÍº ËÒ¤èÒäÁèä´é.
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#9
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¢éÍ 2 ¹Ð¤ÃѺ à¼×èÍäÁèà¢éÒ㨠·ÓµÒÁÇÔ¶Õ¾Õè t.b. ¹Ð¤ÃѺ
2. $\left[\,2+\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ + $\left[\,2-\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ ÁÕ¤èÒà·èҡѺà·èÒäà SoLuTiOn ãËé$\left[\,2+\frac{10}{\sqrt{27} } \right]^\frac{1}{3}=x$áÅÐ$\left[\,2-\frac{10}{\sqrt{27} } \right]^\frac{1}{3}=y$ ¨Ðä´é $x^3+y^3=2+\frac{10}{\sqrt{27}}+2-\frac{10}{\sqrt{27}}=4$ ä´é$x^3+y^3=4=(x+y)(x^2-xy+y^2)$ ·Ó¾¨¹ìËÅѧãËéà»ç¹¡ÓÅѧÊͧÊÑÁºÙóì áµèµéͧ·ÓãËéà»ç¹ºÇ¡¨Ðä´é¤ÅéÒ¡Ѻ¾¨¹ì˹éÒ $x^3+y^3=(x+y)(x^2+2xy+y^2-3xy)$ ·Õèµéͧ $-3xy$ à¾ÃÒÐà´ÔÁÁÕ$-xy$áµèàÃÒ·ÓãËéà»ç¹$+2xy$ à¾×èÍãËé¨Ñ´à»ç¹¡ÓÅѧÊͧÊÑÁºÙóìä´é´Ñ§¹Ñé¹µéͧ$-3xy$ à¾×èÍãËéÁÕ¤èÒà·èÒà´ÔÁ $x^3+y^3=(x+y)((x+y)^2-3xy)$ ¤ÃÒǹÕéàÃÒ¨ÐÁÒËÒ $3xy$ à¾×èÍä»á·¹¤èҹФÃѺ $3xy=3\left[\,2+\frac{10}{\sqrt{27} } \right]^\frac{1}{3} \bullet \left[\,2-\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ $3xy=3\left[\frac{108-100 }{27} \right]^\frac{1}{3}$ $3xy=3\left[\frac{8 }{27} \right]^\frac{1}{3}$ $3xy=3\left[\frac{2^3 }{3^3} \right]^\frac{1}{3}$ $3xy=3\left[\frac{2}{3} \right]$ ´Ñ§¹Ñé¹ $3xy=2$ á·¹¤èÒ$3xy$ ä´é$x^3+y^3=(x+y)((x+y)^2-3xy)=(x+y)((x+y)^2-2)$ ÊÁÁصÔãËé $(x+y)=A$ ä´é$x^3+y^3=A(A^2-2)=4$ $A^3-2A-4=0$¨Ò¡·ÄÉ®ÕàÈÉàËÅ×Í áÅСÒÃËÒÃÊѧà¤ÃÒÐËì¨Ðä´é $(A-2)(A^2+2A+2)=0$ ä´é¤ÓµÍº¤×Í $A=2$ à¾ÃÒо¨¹ìËÅѧäÁèÁդӵͺ·Õèà»ç¹¨Ó¹Ç¹¨ÃÔ§à¾ÃÒÐ $b^2-4ac<0$ $A=x+y=2$ ´Ñ§¹Ñé¹ $\left[\,2+\frac{10}{\sqrt{27} } \right]^\frac{1}{3}$ + $\left[\,2-\frac{10}{\sqrt{27} } \right]^\frac{1}{3}=2$ ËÇѧÇèÒÅÐàÍÕ´¾Í¹Ð¤ÃѺ ¾Í´Õªèǧ¹Õé¼Áà¨Í¤ÃÙ¤³ÔµÈÒʵÃì·Õèà¹é¹ÇԸշӴѧ¹Ñé¹·Ø¡ÍÂèÒ§µéͧµÒÁ¢Ñ鹵͹
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I think you're better than you think you are. 30 ¾ÄÉÀÒ¤Á 2008 19:25 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ RETRORIAN_MATH_PHYSICS |
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