Question

me ajudem nessa questão de equação irracional PF. gab 1
$3x^{2/3} + 2 = 5x^{1/3}$

Answer 1

$3x^{2/3}}+2=5x^{1/3}\\ \\3 \sqrt[3]{x^2}+2=5 \sqrt[3]{x} \ \ \ \ \longrightarrow \ usando\ \ \sqrt[3]{x} =y\\ \\3 (\sqrt[3]{x})^2+2=5 \sqrt[3]{x} \\ \\3 y^2+2=5y\\ \\3 y^2-5y+2=0\\ \\y=\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot3\cdot2}}{2\cdot3}=\frac{5\pm\sqrt{25-24}}{6}=\frac{5\pm\sqrt{1}}{6}=\frac{5\pm1}{6}\\ \\y'=\frac{5+1}{6}=\frac{6}{6}=1\\ y''=\frac{5-1}{6}=\frac{4^{:2}}{6_{:2}}=\frac{2}{3}$

Substituindo a variável auxiliar:
Se y = 1                                               Se y = 2/3
$\sqrt[3]{x} =y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt[3]{x} =y\\ \\\sqrt[3]{x} =1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt[3]{x} =\frac{2}{3}\\ \\(\sqrt[3]{x})^3 =1^3\ \ \ \ \ \ \ \ \ \ (\sqrt[3]{x})^3 =(\frac{2}{3})^3\\ \\x =1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x =\frac{8}{27}$

Prova Real para x = 8/27
$3x^{2/3}}+2=5x^{1/3}\\ \\3 \sqrt[3]{x^2}+2=5 \sqrt[3]{x}\\ \\3 \sqrt[3]{(\frac{8}{27})^2}+2=5 \sqrt[3]{\frac{8}{27}}\\ \\3 \sqrt[3]{[(\frac{2}{3})^3]^2}+2=5 \sqrt[3]{(\frac{2}{3})^3}\\ \\3 \cdot(\frac{2}{3})^2+2=5 \cdot\frac{2}{3}\\ \\3 \cdot\frac{4}{9}+2=\frac{10}{3}\\ \\\frac{4}{3}+2=\frac{10}{3}\\ \\\frac{4}{3}+\frac{6}{3}=\frac{10}{3}\\ \\\frac{10}{3}=\frac{10}{3} \ \ (Ok!)$

Prova Real para x = 1
$3x^{2/3}}+2=5x^{1/3}\\ \\3 \sqrt[3]{x^2}+2=5 \sqrt[3]{x}\\ \\3 \sqrt[3]{1^2}+2=5 \sqrt[3]{1}\\ \\3 \cdot1+2=5 \cdot1\\ \\3+2=5\\ \\5=5 \ \ \ (Ok!)$

$S=\{1,\frac{8}{27}\}$

Answer 2

Boa noite Rafael

essa equação é exponencial e não irracional 

3x^(2/3) + 2 = 5x^(1/3) 

y = x^(1/3) 

3y² - 5y + 2 = 0 

delta
d² = 25 - 24 = 1
d = 1

y1 = (5 + 1)/6 = 1
y2 = (5 - 1)/6 = 2/3 

x^(1/3) = 1

x = 1^3 = 1

x^(1/3) = 2/3 

x = (2/3)^3 = 8/27 

S = (1, 8/27)