SOMA DE RAIZES

a soma das raízes é:


a)20.5
b)30.5
c)10.5
d)23.5
e)33.5

(54x-27)^(1/3)/3=(54x-27/27)^(1/3)=(2x-1)^(1/3)
(1458x-729)^(1/6)=((1458x-729)^(1/3))(1/2)=729^(1/6)((2x-1)^1/3)^1/2)=3((2x-1)^(1/3))^(1/2)

Reescrevendo:
(2x-1)^(1/3)-((2x-1)^(1/3))^(1/2)
Fazendo (2x-1)^(1/3)=y <-> 2x-1≥0 <-> x≥1/2
y-3y^(1/2)=-2
y-3sqrt(y)=-2
y+2=3sqrt(y)
y²+4y+4=9y
y²-5y+4=0
y=1 ou y=4
(2x-1)^(1/3)=1 <-> x1=1
(2x-1)^(1/3)=4 <-> x2=65/2

Somando: x1+x2=67/2=33,5

Espero que seja isso e que te ajude.

" title="\text{Usarei o seguinte fato}\\} \text{se}\,\,a, b, c\,\, \text{sao numeros reais tais que}\,\,a+b+c=0\rightarrow a^3+b^3+c^3=3abc\,\,\text{pode-se provar facilmente!!!}\\\\} \text{dessa forma fazendo}\,\,a=\sqrt[3]{27(2x-1)}=3\sqrt[3]{2x-1}\,\,\,,b=\sqrt[6]{729(2x-1)}=3\sqrt[6]{2x-1}\,\,\text{e}\,\,c=2\,\,\text{veja que}\,\,a+(-b)+c=0\rightarrow a^3+b^3+c^3=3abc\rightarrow \\\\} \rightarrow (\sqrt[3]{2x-1})^3+(-3\sqrt[6]{2x-1})^3+2=(3\sqrt[3]{2x-1}).(-3\sqrt[6]{2x-1}).(2)\,\,\,\text{desenvolvendo...}\\\\} \rightarrow 2x-1+8-27\sqrt{2x-1}=-18\sqrt{2x-1}\rightarrow 2x+7=9\sqrt{2x-1}\,\,\text{quadrando os dois membros...}\\} \rightarrow 4x^2+28x+49=162x-81\rightarrow 4x^2-134x+130=0\\} \rightarrow \text{soma das raizes}=\frac{-b}{a}=\frac{-(-134)}{4}=33,5\, " />

Humpf... por que esses códigos saíram assim ¬¬

dando surgiu o 27 no denominador (54x-27/27)^(1/3)

3=(3³)^1/3=27^(1/3)