#1
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µÃÕ⡳¤ÃѺ
$ 1.¨§ËÒ¤èҢͧ cos42^{\circ} sin168^{\circ} sin306^{\circ} $
$ 1. \frac{-1}{8} 2.\frac{1}{8} 3.\frac{-1}{4} 4.\frac{1}{4}$ $ 2. ¨§ËÒ¤èҢͧ tan20^{\circ} tan40^{\circ} tan80^{\circ} $ $ 3. ¨Ó¹Ç¹¤ÓµÍº¢Í§ ÊÁ¡Òà arccos(cos\theta ) = arcsin(sin6\theta ) , 0 \leqslant \theta $ $ \leqslant \pi $ $ 1. 3 2. 4 3. 5 4. 6 $ $ 4. ¢éͤÇÒÁµèÍ仹Õé¶Ù¡ËÃ×ͼԴ $ $ ¡. ¡Ó˹´ÃÙ»ÊÒÁàËÅÕèÂÁ ABC ÁÕ a,b,c à»ç¹¤ÇÒÁÂÒÇ´éÒ¹µÃ§¢éÒÁÁØÁ A,B,C ¨Ðä´éÇèÒ$ $ \frac{1+cos(A-B)cosC}{1+cos(A-C)cosB} = \frac{a^2+b^2}{a^2+c^2} $ $ ¢. a,b,c \in R , a^2 + b^2 \not= c^2 , a^2 + b^2 \not= 0 $ $ ËÒ¡ A,B \in [ 0,2\pi ) , A \not= B à»ç¹¤ÓµÍº¢Í§ÊÁ¡ÒÃ(µÑÇá»Ã X) acosx + bcosx = c ¨Ðä´éÇèÒ $ $ cos^2(\frac{A-B}{2}) = \frac{c^2}{a^2+b^2}$ $ 5. arcsec\sqrt{x^2+1} + arcsec\sqrt{x^2+2x+2} = arccos(-1) + arctan(-1) $ $ ¼ÅºÇ¡¤ÓµÍºà·èҡѺà·èÒäÃ$ $ 1. -1 2. 1 3. 2 4. 3 $ $ 6. ÊÒÁàËÅÕèÂÁ ABC ÁÕ BC = 10 ¤ÇÒÁÂÒÇÃͺÃÙ»à·èҡѺ 36 ¶éÒ AC=b , AB=c $ $ ¨§ËÒ¤èҢͧ b^2sin2C + c^2sin2B$ $ 1.24\sqrt{bc+144} $ $ 2.48\sqrt{bc+144} $ $ 3.24\sqrt{bc -144}$ $ 4.48\sqrt{bc -144} $ 18 ¡Ñ¹ÂÒ¹ 2011 14:04 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ CHAOS |
#2
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á»Å§·Ø¡¤èÒà»ç¹¿Ñ§¡ìªÑ¹â¤ä«¹ìãËéËÁ´ ¨Ò¡¹Ñ鹨Ѻ¤Ùè áÅéÇãªéÊٵà $2\cos A \cos B = ...$ áÅéÇá·¹¤èÒ $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$
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#3
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¨¢¡·.¶ÒÁ·ÕÅШش´Õ¡ÇèÒäËÁ
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#4
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ÍéÒ§ÍÔ§:
$ = \frac{2-2cos(A+B)cos(A-B)}{2-2cos(A+C)cos(A-C)} $ $ = \frac{2-(cos2A+cos2B)}{2-(cos2A+cos2C)} $ $ = \frac{2-(1-sin^2A+1-sin^2B)}{2-(1-sin^2A+1-sin^2B)} $ $ = \frac{sin^2A+sin^2B}{sin^2A+sin^2C} $ ¨Ò¡ Law of sine ¨Ðä´é $ = \frac{a^2k^2+b^2k^2}{a^2k^2+a^2c^2} $ $ =\frac{a^2+b^2}{a^2+c^2} $ = R.S. |
#5
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¢éÍ 6. ¤Ô´ä´é 4.
ÊÔ觷Õè⨷Âì¶ÒÁ¤×Í 4 à·èҢͧ¾×é¹·ÕèÊÒÁàËÅÕèÂÁ ¾×é¹·ÕèÊÒÁàËÅÕèÂÁà·èҡѺ$\frac{1}{2}bc \sin A $ Åͧ¤Ô´àͧ¡è͹¡çä´é ¨Ò¡Êٵâͧcosine ¡Ñº sine ¨Ðä´é¤èÒ $c\sin B-b \sin C=0$........(1) $c \cos B-b \cos C=\frac{c^2-b^2}{10} $..........(2) ¨Ñº(1)¤Ù³¡Ñº(2) áÅéÇá»Å§ãËé $\sin(B+C)= \sin A$ ÊÔ觷Õè⨷Âì¶ÒÁ $b^2 \sin 2C+c^2 \sin 2B- =2bc \sin A$ áÅéÇà·Õº¡ÑºÊٵþ×é¹·ÕèÊÒÁàËÅÕèÂÁ¢Í§Heron¡ç¨Ðä´é¤ÓµÍº
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"¶éÒàÃÒÅéÁºèÍÂæ ã¹·ÕèÊØ´àÃÒ¨ÐÃÙéÇèÒ¶éÒ¨ÐÅéÁ ÅéÁ·èÒä˹¨Ðà¨çº¹éÍ·ÕèÊØ´ áÅÐÃÙéÍÕ¡ÇèÒµèÍä»·ÓÂѧ䧨ÐäÁèãËéÅéÁÍÕ¡ ´Ñ§¹Ñ鹨§ÍÂèÒ¡ÅÑÇ·Õè¨ÐÅéÁ"...ÍÒ¨ÒÃÂìÍӹǠ¢¹Ñ¹ä·Â ¤ÃÑé§áá㹪ÕÇÔµ·ÕèÊͺ¤³ÔµÊÁÒ¤Á¤³ÔµÈÒʵÃìàÁ×èÍ»Õ2533...¼Áä´éá¤è24¤Ðá¹¹(¨Ò¡ÃéͤÐá¹¹) |
#6
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ÍéÒ§ÍÔ§:
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#7
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¢éÍ 2....·Óä´é 2Ẻ
Ẻ·Õè1 $\tan 20^\circ \tan 40^\circ \tan 80^\circ$ $=\frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ} $ $=\frac{\sin 80^\circ (\cos 20^\circ - \cos 60^\circ)}{\cos 80^\circ (\cos 20^\circ + \cos 60^\circ)} $ $=\frac{2\sin 80^\circ \cos 20^\circ-2\sin 80^\circ \cos 60^\circ}{2\cos 80^\circ \cos 20^\circ+2\cos 80^\circ \cos 60^\circ} $ $=\frac{\sin 100^\circ+\sin 60^\circ-\sin 80^\circ}{\cos 100^\circ+\cos 60^\circ+\cos 80^\circ} $ $\sin 100^\circ =\sin80^\circ,\cos 100^\circ= -\cos 80^\circ$ $=\frac{\sin 60^\circ}{\cos60^\circ} $ $=\tan 60^\circ = \sqrt{3} $ Ẻ·Õè 2....ÁѹÁÕÊÙµÃÇèÒ $\tan (60-x)^\circ \tan x^\circ \tan (60+x)^\circ= \tan 3x^\circ $ á·¹ $x=20^\circ$ $\tan (60-20)^\circ \tan20^\circ \tan (60+20)^\circ= \tan (3\times 20)^\circ$ $\tan 40^\circ \tan 20^\circ \tan 80^\circ = \tan 60^\circ = \sqrt{3} $
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"¶éÒàÃÒÅéÁºèÍÂæ ã¹·ÕèÊØ´àÃÒ¨ÐÃÙéÇèÒ¶éÒ¨ÐÅéÁ ÅéÁ·èÒä˹¨Ðà¨çº¹éÍ·ÕèÊØ´ áÅÐÃÙéÍÕ¡ÇèÒµèÍä»·ÓÂѧ䧨ÐäÁèãËéÅéÁÍÕ¡ ´Ñ§¹Ñ鹨§ÍÂèÒ¡ÅÑÇ·Õè¨ÐÅéÁ"...ÍÒ¨ÒÃÂìÍӹǠ¢¹Ñ¹ä·Â ¤ÃÑé§áá㹪ÕÇÔµ·ÕèÊͺ¤³ÔµÊÁÒ¤Á¤³ÔµÈÒʵÃìàÁ×èÍ»Õ2533...¼Áä´éá¤è24¤Ðá¹¹(¨Ò¡ÃéͤÐá¹¹) |
#8
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ÍéÒ§ÍÔ§:
ãËé $arcsec\sqrt{x^2+1}=A \rightarrow \sec A=\sqrt{x^2+1} ,\cos A=\frac{1}{\sqrt{x^2+1}} ,\sin A=\frac{x}{\sqrt{x^2+1}} $ $arcsec\sqrt{x^2+2x+2}=B \rightarrow \sec B=\sqrt{x^2+2x+2} ,\cos B=\frac{1}{\sqrt{x^2+2x+2}} ,\sin B=\frac{x+1}{\sqrt{x^2+2x+2}}$ $arccos(-1)=C \rightarrow \sin C=0, \cos C=-1$ $arctan(-1)=D \rightarrow \sin D=-\frac{1}{\sqrt{2} } , \cos D=\frac{1}{\sqrt{2} }$ ¨ÑºÊÁ¡ÒÃÁÒãèÊè¿Ñ§¡ìªÑ蹢ͧ cos à¾ÃÒзÕè⨷Âì¡Ó˹´¡ç¤×ÍÁØÁ·Ñé§ÊÕè¤èÒ $\cos(arcsec\sqrt{x^2+1} + arcsec\sqrt{x^2+2x+2})= \cos (A+B)$ $=\frac{1-x-x^2}{\sqrt{(x^2+1)(x^2+2x+2)} } $ $\cos (arccos(-1) + arctan(-1))= \cos (C+D) = -\frac{1}{\sqrt{2} }$ $\frac{1-x-x^2}{\sqrt{(x^2+1)(x^2+2x+2)} }= -\frac{1}{\sqrt{2} }$ $2(x^2+x-1)^2=(x^2+1)(x^2+2x+2)$ $x^4+2x^3-5x^2-6x=0$ $x(x+1)(x+3)(x-2)=0$ $x= 0,-1,-3,2$ ¤èÒ·Õèãªéä´é¤×Í $x=-3,2$ ¼ÅÃÇÁ¢Í§¤ÓµÍºà·èҡѺ $-1$
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"¶éÒàÃÒÅéÁºèÍÂæ ã¹·ÕèÊØ´àÃÒ¨ÐÃÙéÇèÒ¶éÒ¨ÐÅéÁ ÅéÁ·èÒä˹¨Ðà¨çº¹éÍ·ÕèÊØ´ áÅÐÃÙéÍÕ¡ÇèÒµèÍä»·ÓÂѧ䧨ÐäÁèãËéÅéÁÍÕ¡ ´Ñ§¹Ñ鹨§ÍÂèÒ¡ÅÑÇ·Õè¨ÐÅéÁ"...ÍÒ¨ÒÃÂìÍӹǠ¢¹Ñ¹ä·Â ¤ÃÑé§áá㹪ÕÇÔµ·ÕèÊͺ¤³ÔµÊÁÒ¤Á¤³ÔµÈÒʵÃìàÁ×èÍ»Õ2533...¼Áä´éá¤è24¤Ðá¹¹(¨Ò¡ÃéͤÐá¹¹) 22 ¡Ñ¹ÂÒ¹ 2011 14:13 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¡ÔµµÔ |
#9
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#8
$arcsec\sqrt{x^2+1}=A \rightarrow \sec A=\sqrt{x^2+1} ,\cos A=\frac{1}{\sqrt{x^2+1}} ,\sin A=\frac{\left|x\right| }{\sqrt{x^2+1}} $ $arcsec\sqrt{x^2+2x+2}=B \rightarrow \sec B=\sqrt{x^2+2x+2} ,\cos B=\frac{1}{\sqrt{x^2+2x+2}} ,\sin B=\frac{\left|x+1\right|}{\sqrt{x^2+2x+2}}$ $\cos (A+B)=\frac{1-\left|x\right|\left|x+1\right|}{\sqrt{(x^2+1)(x^2+2x+2)} }= -\frac{1}{\sqrt{2} }$ àÇÅÒá¡éÊÁ¡ÒÃäÁè¹èҨЧèÒ ´Ñ§·Õè¤Ø³ËÁÍ·ÓÁÒ¤ÃѺ áµè¤ÓµÍºä´éÍÍ¡ÁÒ¶Ù¡ ÍÒ¨à»ç¹à¾ÃÒС¡ÓÅѧÊͧà¢éÒä»·ÓÅÒ¤èÒÊÑÁºÙóì áµè¨ÃÔ§ æ ¡çäÁèä´é¡¡ÓÅѧÊͧá¤è¤ÃÑé§à´ÕÂÇ áµè¶éÒá»Å§â¨·Âìà»ç¹ $arctan\left|x\right| +arctan\left|x+1\right|=\frac{3\pi}{4} $ ¼ÁÇèÒ¹èÒ¨Ðá¡éÊÁ¡ÒçèÒ¡ÇèÒ¤ÃѺ 22 ¡Ñ¹ÂÒ¹ 2011 16:34 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ lek2554 à˵ؼÅ: ¾ÔÁ¾ì¼Ô´ |
#10
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àËç¹´éǵÒÁ·Õè¾ÕèàÅç¡ÇèÒ ¼ÁÅ×Áàªç¤â´àÁ¹¡Ñºàù¨ì¢Í§ÍÔ¹àÇÍÃìʿѧ¡ìªÑè¹µÃÕ⡳
¼Á¢Íá¡éãªé $\left|\,x\right| =\sqrt{x^2} $ ¡Ñº $\left|\,x+1\right| =\sqrt{(x+1)^2} $ $\frac{1-\sqrt{x^2(x+1)^2}}{\sqrt{(x^2+1)(x^2+2x+2)} }=-\frac{1}{\sqrt{2} } $ $2(1+(x^2(x+1)^2)-2\sqrt{x^2(x+1)^2})=x^4+2x^3+3x^2+2x+2$ $2+2(x^4+2x^3+x^2)-4\left|\,x(x+1)\right|=x^4+2x^3+3x^2+2x+2 $ $x^4+2x^3-x^2-2x= 4\left|\,x(x+1)\right|$ $x(x-1)(x+1)(x+2)=4\left|\,x(x+1)\right|$ ¼ÁäÁè¡¡ÓÅѧÊͧ à¾ÃÒйèҨеԴ¡Ñ¹ÇØè¹ÇÒ ãªé¹ÔÂÒÁ¢Í§¤èÒÊÑÁºÙÃ³ì ´éÇ¡ÒÃá¡à»ç¹ªèǧæ 1.$x>0$ $x(x-1)(x+1)(x+2)=4x(x+1)$ $(x-1)(x+2)=4$ $x^2+x-6=0$ $x=-3,2$ àËÅ×Íá¤è $x=2$ 2.$-1<x<0$ $x(x-1)(x+1)(x+2)=-4x(x+1)$ $(x-1)(x+2)=-4$ $x^2+x+2=0$ ¤èÒ $x$ à»ç¹¨Ó¹Ç¹àªÔ§«é͹ 3.$x<-1$ $x(x-1)(x+1)(x+2)=4x(x+1)$ $(x-1)(x+2)=4$ $x^2+x-6=0$ $x=-3,2$ àËÅ×Íá¤è $x=-3$ ¤èÒ$x=0,-1$....¡ç·ÓãËéÊÁ¡ÒÃ$x(x-1)(x+1)(x+2)=4\left|\,x(x+1)\right|$ à»ç¹¨ÃÔ§ Å᷹ͧ¤èÒ¡ÅѺä»$x=0,-3,2,-1$ $\frac{1-\sqrt{x^2(x+1)^2}}{\sqrt{(x^2+1)(x^2+2x+2)} }=-\frac{1}{\sqrt{2} } $ ÁÕ¤èÒ·Õèà»ç¹¨ÃÔ§¤×Í $x=2,-3$
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#11
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¼Á·ÓẺ¹Õé¤ÃѺ
$arcsec\sqrt{x^2+1} + arcsec\sqrt{x^2+2x+2} = arccos(-1) + arctan(-1) $ $arctan\left|x\right| +arctan\left|x+1\right|=\pi -\dfrac{\pi}{4}=\dfrac{3\pi}{4} $ $tan(arctan\left|x\right| +arctan\left|x+1\right|)=tan\frac{3\pi}{4} $ $\dfrac{|x|+ |x+1|}{1-|x||x+1|} =-1$ $|x|+ |x+1|-|x||x+1|+1=0$ $x<-1;\quad\quad-x-x-1-x(x+1)+1=0\rightarrow x^2+3x=0\rightarrow x=-3,0\rightarrow x=-3$ $-1\leqslant x\leqslant 0;-x+x+1+x(x+1)+1=0\rightarrow x^2+x+2=0\rightarrow x\in \phi $ $x> 0;\quad\quad\quad x+x+1-x(x+1)+1=0\rightarrow x^2-x-2=0\rightarrow x=-1,2\rightarrow x=2$ $\therefore x=-3,2$ µÃǨÊͺ¤ÓµÍºáÅéÇãªéä´é·Ñé§ 2 ¤èÒ |
#12
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ÇԸբͧ¾ÕèàÅ硧èÒ¡ÇèÒ¤ÃѺ....
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