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Re: [obm-l] Questão
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] Questão
- From: Marcos Martinelli <mffmartinelli@xxxxxxxxx>
- Date: Mon, 12 Dec 2005 14:16:53 -0200
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Se a+b+c=0 -> a+b=-c -> a^2+2ab+b^2=c^2. Como a^2+b^2=1-c^2 -> 1-c^2+2ab=c^2
-> c^2=(1+2ab)/2. Mas (a^2+b^2)^2=(1-c^2)^2 -> a^4+b^4+2a^2b^2=[(1-2ab)/2]^2 ->
a^4+b^4+2(ab)^2=[1-4ab+4(ab)^2]/4 -> a^4+b^4+2(ab)^2=1/4-ab+(ab)^2. Mas ainda temos que ab=c^2-1/2 -> a^4+b^4+(c^2-1/2)^2=1/4-(c^2-1/2). E então finalmente:
a^4+b^4+c^4-c^2+1/4=1/4-c^2+1/2 -> a^4+b^4+c^4=1/2.