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[˹ѧÊ×Í ÊÍǹ.] ¨Ó¹Ç¹àªÔ§«é͹
1. ¨§¾ÔÊÙ¨¹ìÇèÒ¶éÒ $z^5=1$ áÅéÇ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2$
2. ¨§ËҼźǡ¢Í§ $\displaystyle \binom{n}{1}-3\binom{n}{3}+3^2\binom{n}{5}-3^3\binom{n}{7}+...$ 3. ãËé $a,b,c,d$ à»ç¹¨Ó¹Ç¹àµçÁºÇ¡ ¨§ËÒ $a+b+c+d$ ¶éҼźǡ¢Í§¤ÇÒÁÂÒÇ·Ø¡´éÒ¹áÅÐàÊé¹·á§ÁØÁ¢Í§ÃÙ» $12$ àËÅÕèÂÁ´éÒ¹à·èÒṺã¹Ç§¡ÅÁÃÑÈÁÕ $12$ ˹èÇ ÍÂÙèã¹ÃÙ» $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ 4.(Íѹ¹ÕéäÁèà¡ÕèÂǡѺàªÔ§«é͹) ¨§áÊ´§ÇèÒ¨ÐäÁèÁÕ¾ËعÒÁ $P(x)$ ·ÕèÁÕÊÑÁ»ÃÐÊÔ·¸Ôìà»ç¹¨Ó¹Ç¹àµçÁ«Öè§ $P(10)+P(10^2)+P(10^3)+...+P(10^9)=10^{10}$ 5.¨§ËÒ·Ø¡¤ÙèÍѹ´Ñº¤ÓµÍº¢Í§ÊÁ¡Òà (¶éÒ $\in C$ ÃÇÁ´éÇÂ) ¢Í§ $y^4+y^3+y^2+y=x^2+x$ |
#2
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àÍ...¶éÒàÍÒÊͧà·ÍÁááÁҺǡ¡Ñ¹¨Ðä´éÍÐäùÐ
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site:mathcenter.net ¤Ó¤é¹ |
#3
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ÍéÒ§ÍÔ§:
$(1+x)^n = \binom{n}{0} + \binom{n}{1}x+\binom{n}{2}x^2+\cdots$ ¨Ðµé᷹ͧ $x$ à»ç¹à·èÒäËÃè¹Ð¨Ö§¨Ðä´éÍÐäÃẺ¹Õé $\binom{n}{0} - \binom{n}{1}x +\binom{n}{2}x^2+\cdots$ ¶éÒàÍÒÊͧÍѹ¹ÕéÁÒź¡Ñ¹à·ÍÁ·Õèà»ç¹àÅ¢¤Ùè¨ÐËÒÂä»ËÁ´àÅÂáÎÐ àÍ..áÅéǨеé᷹ͧ $x$ à»ç¹à·èÒäËÃèÍÕ¡¹Ð¨Ö§¨Ðä´éÊÔ觷Õè⨷Âìµéͧ¡ÒÃ
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site:mathcenter.net ¤Ó¤é¹ |
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4. $P(10) \equiv P(10^2) \equiv \cdots \equiv P(10^9) \pmod 9$
$9 \mid (P(10)+P(10^2)+\cdots + P(10^9))$ ¨Ö§à»ç¹ $10^{10}$ äÁèä´é¤ÃѺ
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#5
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¢é͹Õé¹ÕèËҨӹǹàµçÁ·Ñé§ËÁ´àËÃͤÃѺ
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site:mathcenter.net ¤Ó¤é¹ |
#6
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¤×͵͹¹Õéä´é $y$ ä´é $3$ µÑÇ ¤×Í $-1,0,1$ (ÁÒ¨Ò¡ÇÔªÒ·Äɮըӹǹ)
áµèÍÂÒ¡·ÃÒº·Ø¡¤ÓµÍºã¹¨Ó¹Ç¹¨ÃÔ§ ËÃ×Í㹨ӹǹàªÔ§«é͹¹Ð¤ÃѺ |
#7
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ÍéÒ§ÍÔ§:
¨Ó¹Ç¹¨ÃÔ§¡çÁÕ͹ѹµìàËÁ×͹¡Ñ¹ áµèµéͧ¤Ô´·Õè x,y µéͧäÁèà»ç¹¨Ó¹Ç¹àªÔ§«é͹´éÇ |
#8
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[quote=¿Ô¹Ô¡«ìàËÔ¹¿éÒ;172564]1. ¨§¾ÔÊÙ¨¹ìÇèÒ¶éÒ $z^5=1$ áÅéÇ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2$
1) ¨Ò¡ $\dfrac{z}{1+z^2}=\dfrac{z\times z^{3}}{(1+z^2)\times z^{3}}=\dfrac{z^4}{1+z^3}$ 2) ¨Ò¡ $\dfrac{z^2}{1+z^4}=\dfrac{(z^2)\times z}{(1+z^4)\times z}=\dfrac{z^3}{1+z}$ 3) $\therefore\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2(\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4})$ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2(\dfrac{z(1+z^4)+(1+z^2)(z^2)}{(1+z^2)(1+z^4)})$ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2(\dfrac{z+z^5+z^2+z^4}{1+z^2+z^4+z^6})$ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2(\dfrac{1+z+z^2+z^4}{1+z+z^2+z^4})$ $\dfrac{z}{1+z^2}+\dfrac{z^2}{1+z^4}+\dfrac{z^3}{1+z}+\dfrac{z^4}{1+z^3}=2(1)=2$ |
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