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#1
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͹ءÃÁ àÅç¡æ¤ÃѺ
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áµè¼Á¤Ô´äÁèÍÍ¡¨ÃÔ§æ ¢Í Guide ˹èͤÃѺ ËÃ×Í«è͹ Solution ¡çä´é ãËé A,B,C,D,E,F à»ç¹ÅӴѺàÅ¢¤³Ôµ â´Â A äÁèà·èҡѺ B ¨§ËÒ¤èÒ x,y ¨Ò¡ÃкºÊÁ¡Òà AX+BY = C DX+EY = F IF 1/(y+z) + 1/(z+x) + 1/(x+y) à»ç¹ÅӴѺàÅ¢¤³ÔµáÅéÇ ¨§¾ÔÊÙ¨¹ìÇèÒ $x^2,y^2,z^2$ à»ç¹ÅӴѺàÅ¢¤³Ôµ ¢Íâ·ÉàÃ×èͧ Latex ´éǹФÃѺ¾Í´Õà¹çµªéÒáÅéÇÁѹäÁè¢Öé¹ |
#2
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¢éÍáá á¡éÊÁ¡Òû¡µÔ¤ÃѺ ãËéÊѧࡵÅӴѺàÅ¢¤³Ôµ
¢éÍÊͧ äÁèà¢éÒã¨â¨·Âì¤ÃѺ ⨷ÂìºÍ¡ÅӴѺä˹¤ÃѺ |
#3
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¢éÍ1
$AX+BY = C$......(1) $DX+EY = F$.......(2) $AEX+BEY=CE$ $BDX+BEY=BF$ $x=\frac{CE-BF}{AE-BD} $ $A,B,C,D,E,F$à»ç¹ÅӴѺàÅ¢¤³Ôµ $F-E=E-D=D-C=C-B=B-A=m$ $x=\frac{CE-BF}{AE-BD}=\frac{(A+2m)(A+4m)-(A+m)(A+5m)}{A(A+4m)-(A+m)(A+3m)} $ $=\frac{3m^2}{-3m^2} $ $=-1 $ $Y=\frac{C+A}{B}= \frac{2A+2m}{A+m} =2$
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#4
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¢éÍ2 à¢éÒã¨ÇèÒ⨷Âì¹èÒ¨Ðà»ç¹
$\frac{1}{(y+z)} , \frac{1}{(z+x)} ,\frac{1}{(x+y)} $ à»ç¹ÅӴѺàÅ¢¤³ÔµáÅéÇ ¨§¾ÔÊÙ¨¹ìÇèÒ $x^2,y^2,z^2$ à»ç¹ÅӴѺàÅ¢¤³Ôµ àÃÒ¨Ðä´éÇèÒ$\quad 2\left(\, \frac{1}{(z+x)}\right)= \frac{1}{(y+z)} +\frac{1}{(x+y)}$ $\quad 2\left(\, \frac{1}{(z+x)}\right)= \frac{(x+y)+(y+z)}{(x+y)(y+z)} $ $\quad 2\left(\, (x+y)(y+z)\right)= (z+x)\left\{\,(x+y)+(y+z)\right\} $ $\quad 2\left(\, (xy+yz+xz+y^2)\right)=\left\{\,(z+x)(x+y)+(z+x)(y+z)\right\} $ $\quad 2\left(\, (xy+yz+xz+y^2)\right)=\left\{\,(xz+x^2+xy+zy)+(zy+xy+z^2+xz)\right\} $ $\quad 2y^2=x^2+z^2 \rightarrow y^2=\frac{x^2+z^2}{2} $ ËÃ×Íà¢Õ¹à»ç¹$y^2-x^2=z^2-y^2$ ä´éµÒÁ⨷Âìµéͧ¡ÒÃ
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#5
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¢ÍàÊÃÔÁ¢éÍ1ÍÕ¡ÇÔ¸Õ¹èФÃѺ ¨Ò¡ÅӴѺàÅ¢¤³Ôµ¨Ðä´é
Ô B-A=C-B 2B-A=C ¹ÓÁÒà·Õº¡Ñº Ax+By=C ¨Ðä´é x=-1,y=2 ¡çä´éàËÁ×͹¡Ñ¹¤ÃѺ |
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