Algebra - probleme mai dificile

Numere reale. Radicali. Calcul algebric. Ecuatii si inecuatii. Patrulatere. Arii. Asemanarea triunghiurilor. Relatii metrice in triunghiul dreptunghic. Cercul.
Cris_n
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Membru din: 22 Oct 2016, 13:21

Algebra - probleme mai dificile

Mesaj de Cris_n » 22 Oct 2016, 13:23

Ma poate ajuta cineva?
1.
rezolvati in multimea nr rationale
(x-1)/2014 + (x-3)/2012 + (x-5)/2010 + ....+ (x-2011)/4 + (x-2013)/2 = (x-2014)/1 + (x-2012)/3 + (x-2010)/5 +.... + (x-4)/2011 + (x-2)/2013

2.rezolvati in Q
A) x*3^2004 =(3^2004 -1) : (1+1/3+1/3^2 +....+1/3^2003 )
B) x(1-1/2)(1-1/3)(1-1/4)......(1-1/2011)=1/2012
C) x(-1/(1*2)-1/(2*3)-1/(3*4)-.......1/(2011*2012)=-2011/2012

3. Rezolvati in Multimea nr rationale
(2x+1)/2 + (2x+2)/3 + (2x+3)/4+.....+[2x+(n-1)]/n +(2x+n)/(n+1)= n


Multumesc frumos[/code]

DD
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Membru din: 06 Aug 2010, 17:59

Mesaj de DD » 22 Oct 2016, 18:58

1)vom restrange expresia sub o forma mai usor de manevrat.
X*∑_(k=1)^1007▒〖1/(2016-2k)-∑_(k=1)^1007▒(2k-1)/(2016-2k)=〗X*∑_(k=1)^1007▒1/(2015-2k)—∑_(k=1)^1007▒2k/(2015-2k) sau; X*∑_(k=1)^1007▒(1/(2015-2k)-1/(2016-2k)) =∑_(k=1)^1007▒(-(2k-1)/(2016-2k)+2k/(2015-2k)) = ∑_(k=1)^1007▒2k(1/(2015-2k)-1/(2016-2k)) +∑_(k=1)^1007▒(1/(2016-2k)) de unde
X=(∑_(k=1)^1007▒2k(1/(2015-2k)-1/(2016-2k)) +∑_(k=1)^1007▒(1/(2016-2k)) )/(∑_(k=1)^1007▒(1/(2015-2k)-1/(2016-2k)) )=(∑_(k=1)^1007▒(2k/(2016-2k)(2015-2k) +1/(2016-2k)) )/(∑_(k=1)^1007▒(1/(2016-2k)(2015-2k) ) )
Imi pare f rau pentru modul de scriere dar cu putin efort se poate intelege

DD
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Membru din: 06 Aug 2010, 17:59

Mesaj de DD » 22 Oct 2016, 19:36

X=(1-1/3^2004)/(1+1/3+1/3^2+..+1/3^2003)=(1-1/3^2004)^2/(1-1/3)=(3/2)*(1-1/3^2004)^2
X*∏_(K=2)^2011▒〖(1-1/K)=X*1/2*2/3*3/4*..*2010/2011=X/2011〗=1/2012->X=2011/2012
X*∑_(K=1)^2011▒(-1/(K(K+1)) = X* X*∑_(K=1)^2011▒(1/(K+1)-1/K) =X*[1/2-1/1+1/3-1/2+1/4-1/3..+1/2012-1/2011]=X*[-1+1/2012]=-2011/2012->X=1

DD
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Membru din: 06 Aug 2010, 17:59

Mesaj de DD » 22 Oct 2016, 19:53

2X∑_(K=1)^N▒(1/(K+1)) +∑_(K=1)^N▒(K/(K+1)) = N→2X*∑_(K=1)^N▒(1/(K+1)) =N-∑_(K=1)^N▒(K/(K+1)) =∑_(K=1)^N▒(1-K/(K+1)) =∑_(K=1)^N▒(1/(K+1)) →X=1/2

Cris_n
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Membru din: 22 Oct 2016, 13:21

Mesaj de Cris_n » 22 Oct 2016, 20:20

Multumesc frumos

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