Question

demonstre que 5^0,25>2^1/3> \sqrt[6]{3} [/tex]

Answer 1

$5^{0,25} \ \textgreater \ 2^{ \frac{1}{3} } \ \textgreater \ \sqrt[6]{3}$

0,25 = 25/100 = 1/4 

$5^{ \frac{1}{4} } \ \textgreater \ 2^{ \frac{1}{3} } \ \textgreater \ \sqrt[6]{3}$

$\sqrt[4]{5} \ \textgreater \ \sqrt[3]{2} \ \textgreater \ \sqrt[6]{3}$ 
igualando os índices --> mmc(4,3,6) = 12

$\sqrt[12]{5^3} \ \textgreater \ \sqrt[12]{2^4} \ \textgreater \ \sqrt[12]{3^2}$

$\sqrt[12]{125} \ \textgreater \ \sqrt[12]{16} \ \textgreater \ \sqrt[12]{9}$