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ÊÁѤÃÊÁÒªÔ¡ | ¤ÙèÁ×Í¡ÒÃãªé | ÃÒª×èÍÊÁÒªÔ¡ | »¯Ô·Ô¹ | ¢éͤÇÒÁÇѹ¹Õé | ¤é¹ËÒ |
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à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
#1
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¢éÍÊͺàŢ¡¡ÓÅѧ
ãËé $a > 0$ ¤èҢͧ $(\frac{\sqrt[4]{a^3} - \sqrt[4]{a}}{1-\sqrt{a}} + \frac{1+\sqrt{a} }{\sqrt[4]{a}})^2 (1 + \frac{2}{\sqrt{a}} + \frac{1}{a})^\frac{-1}{2} $ ¤×Í¢éÍã´
¶éÒ $\sqrt{a-6} + \sqrt{a+6} = \frac{12}{\sqrt{a-16}}$ áÅéÇ $\sqrt{a-1} - \sqrt{a-9}$ ¤×Í¢éÍã´ |
#2
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¢éÍáá¼Áä´é $\frac{a}{2(\sqrt{a}+1 )}$ ÍФÃѺ
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#3
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ÍéÒ§ÍÔ§:
$ = \left(\dfrac{\sqrt[4]{a} \cdot \sqrt[4]{a^3 } -\sqrt[4]{a} \cdot \sqrt[4]{a} + (1-a) }{\sqrt[4]{a} (1-\sqrt{a} )} \right)^2 \left(\dfrac{1}{\sqrt{1 + \dfrac{2}{\sqrt{a}} + \dfrac{1}{a}}} \right)$ $ = \left( \dfrac{a-\sqrt{a} +1 - a }{ \sqrt[4]{a}(1-\sqrt{a}) } \right)^2 \left( \dfrac{1}{\sqrt{(1+ \frac{1}{\sqrt{a} })^2} } \right)$ $ = \left(\dfrac{1-\sqrt{a} }{\sqrt[4]{a}(1-\sqrt{a} ) } \right)^2 \left(\dfrac{1}{1+ \frac{1}{\sqrt{a} }} \right)$ $ = \dfrac{1}{\sqrt{a} } \cdot \dfrac{1}{1+ \frac{1}{\sqrt{a} }}$ $ = \dfrac{1}{\sqrt{a} +1 }$
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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´) |
#4
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ÍèÒÇ ¼Ô´«Ð§Ñé¹
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#5
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ÍéÒ§ÍÔ§:
ààµè¢é͹Õé ´ÙàËÁ×͹¨ÐäÁèÁդӵͺà»ç¹¨Ó¹Ç¹¨ÃÔ§
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Vouloir c'est pouvoir |
#6
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#5
ÁÕ Real Root ¹Ð¤ÃѺ |
#7
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$\sqrt{a-6}\ge 0$ $\rightarrow a\ge 6$
$$\frac{12}{\sqrt{a-16}}=\sqrt{a-6} + \sqrt{a+6} \ge \sqrt{12}$$ $$\Leftrightarrow \sqrt{a-16}\leq \sqrt{12}\rightarrow a<6$$ à¡Ô´¢é͢ѴààÂé§
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Vouloir c'est pouvoir |
#8
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ºÃ÷ѴÊØ´·éÒÂÁÒÍÂèÒ§ääÃѺ
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#9
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$\rightarrow 12\ge \sqrt{12}\sqrt{a^2-16}\rightarrow \sqrt{a^2-16}\leq \sqrt{12}$
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Vouloir c'est pouvoir |
#10
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¤ÓµÍºàÅ¢äÁèÊÇ Êèǹ trick ¢Í§¢é͹Õé ãËéà¾è§¡ÊÔ³·Õè $(a+6)-(a-6) = 12$ áÅéÇá¡éÊÁ¡ÒøÃÃÁ´Ò ¡çÍÍ¡áÅéǤÃѺ
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#11
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#9
ËÁÒ¶֧·ÓäÁ $a<6$ |
#12
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¢Íâ·É·Õ¤ÃѺ ¼ÁàºÅÍàͧ¹Ö¡ÇèÒÁÕÃÙ·ÍÕ¡ 555+
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Vouloir c'est pouvoir |
#13
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ÍéÒ§ÍÔ§:
¤Ù³´ÑÇ $\sqrt{a+6}-\sqrt{a-6}$ ·Ñé§Êͧ¢éͧ $(a+6)-(a-6)=\dfrac{12}{\sqrt{a-16}} \cdot \sqrt{a+6}-\sqrt{a-6}$ $\sqrt{a-16}=\sqrt{a+6}-\sqrt{a-6}$ $a-16=2a-2\sqrt{a^2-36}$ $2\sqrt{a^2-36}=a+16$ $4a^2-144=a^2+32a+256$ $3a^2-32a-400=0$ $a=\dfrac{32\pm\sqrt{1024+4800}}{6}=\dfrac{32\pm 8\sqrt{91}}{6}$ ä´é¤èÒ a à»ç¹áºº¹Õé¶Ù¡äËÁ¤ÃѺ
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no pain no gain |
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