#1
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ú¡Ç¹ªèÇÂà©ÅÂ
ú¡Ç¹ªèÇÂà©ÅÂ+áÊ´§ÇÔ¸Õ·Ó·Õ¤èÐ (ÃÍ·´ªèÒ§Áѹ )
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#2
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·Óä´é·Ø¡¢éÍàŹРà©Å¢éÍ·Õè¤Ø³ÂѧäÁèä´é·ÓÅСѹ
$\frac{a}{4-a}=k,\frac{b}{7-b}=k,\frac{c}{13-c}=k$ $a+b+c =16 \therefore 4k-ak+7k-bk+13k-ck=16$ $24k-(a+b+c)k =16$ $24k-16k =16$ $k = 2$ $á·¹¤èÒ¨Ðä´é a=\frac{8}{3},b=\frac{14}{3},c=\frac{26}{3}$ $c-b-a=\frac{4}{3}$ |
#3
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¢éÍ·Õèà¢Õ¹¤ÓµÍºÊÕá´§¡ç·ÓäÁèä´é¤èÐ
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#4
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¢éÍ 7 ºÇ¡·Ñé§ÊÁ¡ÒôéÇ 2 ¨Ðä´é
$$\frac{a+b+c}{c} =\frac{a+b+c}{b} =\frac{a+b+c}{b}$$ áºè§à»ç¹ 2 ¡Ã³Õ ¤×Í $$¡Ã³Õ a + b + c = 0 ¨Ðä´é$$ $$a+b = -c , a+c=-b,b+c=-a $$ ¹Óä»á·¹¤èÒ¨Ðä´é x = -1 $$¡Ã³Õa+b+c\not= 0$$ ËÒõÅÍ´´éÇ a+b+c ¨Ðä´é $$\frac{1}{c} =\frac{1}{b} =\frac{1}{a} $$ ËÃ×Í a = b = c ¹Óä»á·¹ ¨Ðä´é x = 8 ´Ñ§¹Ñé¹ x = -1,8 |
#5
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¢éÍ4. ãËéËÒ¤èÒa,b áÅÐ c «Öè§ä´é $a=1, b=\frac{3\pm \sqrt{13}}{2} áÅÐ c=2\pm \sqrt{3} $
áÅéÇá·¹¤èÒ¨Ðä´é $3(a^3+\frac{1}{a^3})=3(1+1)=6$ $b^3-\frac{1}{b^3}=(b-\frac{1}{b})(b^2+1+\frac{1}{b^2})=3(11+1)=36$ $c^3+\frac{1}{c^3}=(c+\frac{1}{c^3})(c^2-1+\frac{1}{c^2})=4(14-1)=52$ ¤ÓµÍº$=6-36+52=22$ |
#6
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ÍéÒ§ÍÔ§:
$a+\dfrac{1}{a}=2$ $b-\dfrac{1}{b}=3$ $c+\dfrac{1}{c}=4$ ÅͧµèʹٹФÃѺ |
#7
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àÍèÍ...ÊÒÁµÑǹÕéÁѹÁÒä´éä§ÍФÐ
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#8
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¢éÍ 1,3 ¼ÁÂѧäÁèá¹è㨤ӵͺ ÃÍ·èÒ¹Í×è¹ÁÒ¤Ô´ãËé¹Ð¤ÃѺ
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#9
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¨Ò¡àÍ¡Åѡɳì¹Õé
$\sqrt{a} +\sqrt{b} = \sqrt{a+b+2\sqrt{ab} } $ ½Ñ觫éÒ¨Ðä´é $\sqrt{2}-\sqrt{1}+x$ ½Ñ觢ÇÒ $ \sqrt{\sqrt{5}+2 }+\sqrt{\sqrt{5}-2 } = \sqrt{\sqrt{5}+2 + \sqrt{5}-2 +\sqrt{1 }} $ áÅéǽÑ觢ÇҤس´éÇ $\frac{\sqrt{\sqrt{5}-1 } }{\sqrt{\sqrt{5}-1 } } $ (¤Ø³à¾×èÍãËéµÑÇÊèǹËÒÂ) ¤§·ÓµèÍä´éáÅéÇãªèäËÁ¤ÃѺ §§»Ð¤ÃѺ 555 ¢éÍ 3 ¤ÅéÒÂæ¡Ñ¹ ¢éÍ 3 µÍº $9-4\sqrt{2}$ 30 àÁÉÒ¹ 2013 01:18 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 8 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¤¹ÍÂÒ¡à¡è§ |
#10
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¨Ò¡ÊÁ¡ÒÃáá
$a^2+1=2a$ (ÂéÒ¢éÒ§) ËÒà a ·Ñé§ÊÁ¡Òà ¡ç¨Ðä´éàͧ¤ÃѺ b c ¡ç·ÓàËÁ×͹ a ¤ÃѺ |
#11
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ÍéÒ§ÍÔ§:
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