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à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
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#31
09 µØÅÒ¤Á 2010, 17:11
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ÍéÒ§ÍÔ§:
¨Ðä´éÇèÒ $x^2+y^2$ ÁÕ¤èÒÊÙ§ÊØ´àÁ×èÍ $r=DO$ µÒÁÃÙ»¤ÃѺ ´Ñ§¹Ñé¹ $x^2+y^2$ ÁÒ¡ÊØ´ $= 27^2$ ËÁÒÂà赯 ÇÔ¸Õ¡Ò÷Ӣͧ¤Ø³ gon àÃÒÊÒÁÒöÊÁÁµÔãËé $x=-5+14cos\Theta$ áÅÐ $y=12+14sin\Theta $ ÁÒÂÑ§ä§ ¼ÁäÁèÁÕ¤ÇÒÁÃÙé¨ÃÔ§ æ ú¡Ç¹·èÒ¹¼ÙéÃÙéªèÇÂ͸ԺÒÂ˹èÍÂä´éäËÁê¤ÃѺ 
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#33
09 µØÅÒ¤Á 2010, 22:22
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ÍéÒ§ÍÔ§:
¨Ò¡ÊÁ¡ÒÃ⨷Âì¨Ðä´é $x^2+10x+y^2-24y-27=0$ ¨Ò¡¤ÇÒÁÊÑÁ¾Ñ¹¸ì¹Õéµéͧ¨Ñ´ÃÙ»ÊÁ¡ÒÃãËÁèà¾×èÍãËéãªéà§×è͹ä¢â¨·Âìä´é ËÅѧ¨Ò¡¼èÒ¹ÍÊÁ¡ÒÃ⤪Õä»áÅéÇ «Öè§ÊÒÁÒö¨Ñ´ÍÍ¡ÁÒà»ç¹áºº¹Õéä´é $\dfrac{x^2+y^2+27}{2}=x(x+5)+y(y-12)$ $~~~~~~~~~~~~~~~~\leq \sqrt{(x^2+y^2)((x+5)^2+(y-12)^2)}$ $~~~~~~~~~~~~~~~~=14\sqrt{x^2+y^2}$ ´Ñ§¹Ñé¹ $x^2+y^2+27\leq 28\sqrt{x^2+y^2}$ $(\sqrt{x^2+y^2}-1)(\sqrt{x^2+y^2}-27)\leq 0$ $1\leq \sqrt{x^2+y^2}\leq 27$
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site:mathcenter.net ¤Ó¤é¹
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#35
09 µØÅÒ¤Á 2010, 23:20
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ÍéÒ§ÍÔ§:
 ÇÔ¸Õ·ÓÊǧÒÁ·Õà´ÕÂǤÃѺ ¤Ø³ nooonuii 
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àÇÅÒ·ÕèàËÅ×ÍÍÂÙèÁÕÇÔ¸Õ¡ÒÃãªéÊͧẺ ¤×Í ·Ò§·ÕèàÃÕº§èÒÂäÁèÁÕÍÐäà ¡ÑºÍÕ¡·Ò§ ·Õè·Ø¡ÍÂèÒ§ÅéǹÁËÑȨÃÃÂì 
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#36
10 µØÅÒ¤Á 2010, 18:42
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ÍéÒ§ÍÔ§:
¨Ðä´é $x^2-4y^2=4$ ËÃ×Í $|x|^2-4|y|^2=4$ ......(1) ãËé $|x|-|y| = c$ ......(2) á¡éÊÁ¡ÒèÐä´é $|x|= \frac{8c\pm \sqrt{16c^2-48} }{6}$ ¡ÃÒ¿ÁըشÃèÇÁ¡Ñ¹ $ \therefore 16c^2-48\geqslant 0$ ¨Ðä´é $c\leqslant -\sqrt{3}$ ËÃ×Í $c\geqslant \sqrt{3} $ áµè $|x|\geqslant0 $ ´Ñ§¹Ñé¹ $c\geqslant \sqrt{3} $ $\therefore |x|-|y| \geqslant \sqrt{3}$ ÊÃØ»ä´éÇèÒ $|x|-|y|$ ÁÕ¤èÒµèÓÊØ´à·èҡѺ $\sqrt{3}$ 10 µØÅÒ¤Á 2010 18:44 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ lek2554
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#39
12 µØÅÒ¤Á 2010, 20:19
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3. ãËé $x\in[-\frac{5\pi}{12},-\frac{\pi}{3}]$ ¨§ËÒ¤èÒÊÙ§ÊØ´¢Í§ $tan(x+\frac{2\pi}{3})-tan(x+\frac{\pi}{6})+cos(x+\frac{\pi}{6})$
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àÇÅÒ·ÕèàËÅ×ÍÍÂÙèÁÕÇÔ¸Õ¡ÒÃãªéÊͧẺ ¤×Í ·Ò§·ÕèàÃÕº§èÒÂäÁèÁÕÍÐäà ¡ÑºÍÕ¡·Ò§ ·Õè·Ø¡ÍÂèÒ§ÅéǹÁËÑȨÃÃÂì
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#40
14 ¾ÄȨԡÒ¹ 2010, 18:16
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