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#1
20 àÁÉÒ¹ 2015, 08:08
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µÃÕ⡳
¡Ó˹´ $a,b >0$ áÅÐ $\dfrac{sin^4x}{a}+ \dfrac{cos^4x}{b}= \dfrac{1}{a+b}$
¨§ËÒ $\dfrac{sin^{2012}x}{a^{1005}}+\dfrac{cos^{2012}x}{b^{1005}}$ 
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#3
20 àÁÉÒ¹ 2015, 19:02
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¤Ô´¨Ò¡¢éÍÁÙÅ⨷Âì·ÕèãËéÁÒä´éÇèÒ
$sin^2x=\frac {a}{a+b} $ áÅÐ $cos^2x=\frac {b}{a+b} $ áÅéǹÓä»á·¹¤èÒ㹤ӶÒÁ¨Ðä´é $(a+b)^{-1005}$ 20 àÁÉÒ¹ 2015 19:04 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ artty60
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#4
20 àÁÉÒ¹ 2015, 20:03
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Åͧãªé ÍÊÁ¡Òà cauchy schwarz ´Ù¤ÃѺ
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#5
20 àÁÉÒ¹ 2015, 21:51
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#4
 $\dfrac{sin^4x}{a}+\dfrac{cos^4x}{b} \ge \dfrac{(sin^2x+cos^2x)^2}{a+b}=\dfrac{1}{a+b}$ Equality holds if and only if$\quad \dfrac{sin^2x}{a}=\dfrac{cos^2x}{b}=k$
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Hope is what makes us strong. It's why we are here. It is what we fight with when all else is lost. 20 àÁÉÒ¹ 2015 21:52 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ FranceZii Siriseth
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