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Jhonyudsonbr
@Jhonyudsonbr
January 2020
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Sabendo que 3x-1 é fator de : 12x³-19x²+8x-1 então as soluções reais das equação:
12.(3^3x)-19.(3^2x)+8.(3^x)-1=0 somam:
resolução por favor !!
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albertrieben
Boa tarde Jhonyduson
12*3^3x - 19*3^2x + 8*3^x - 1 = 0
y = 3^x
12y^3 - 19y^2 + 8y - 1 = 0
3^x - 1 = y - 1
(12y^3 - 19y^2 + 8y - 1)/(y - 1) = 12y^2 - 7y + 1
12y^2 - 7y + 1 = 0
delta
d^2 = 49 - 48 = 1
d = 1
y2 = (7 + 1)/24 = 8/24 = 1/3
y3 = (7 - 1)/24 = 6/24 = 1/4
y1 = 1
3^x1 = 1 = 3^0 , x1 = 0
3^x2 = 1/3 = 3^-1 , x2 = -1
3^x3 = 1/4 , x3 = -2log(2)/(log(3)) = -1.2619
soma
S = 0 - 1 - 1.2619 = -2.2619
2 votes
Thanks 1
jhonyudsonbr
bem detalhado, obrigado mesmo !!
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12*3^3x - 19*3^2x + 8*3^x - 1 = 0
y = 3^x
12y^3 - 19y^2 + 8y - 1 = 0
3^x - 1 = y - 1
(12y^3 - 19y^2 + 8y - 1)/(y - 1) = 12y^2 - 7y + 1
12y^2 - 7y + 1 = 0
delta
d^2 = 49 - 48 = 1
d = 1
y2 = (7 + 1)/24 = 8/24 = 1/3
y3 = (7 - 1)/24 = 6/24 = 1/4
y1 = 1
3^x1 = 1 = 3^0 , x1 = 0
3^x2 = 1/3 = 3^-1 , x2 = -1
3^x3 = 1/4 , x3 = -2log(2)/(log(3)) = -1.2619
soma
S = 0 - 1 - 1.2619 = -2.2619