Questão Potência com Expoente Fracionário! :D

por ismael1008,3 Ter 12 Jul 2016, 11:29

(UNIFOR - 2003) Simplificando-se

(2^-2/3 - 3^-2/3) . (raiz cúbica de 3 - raiz cúbica de 2)^-1 . raiz cúbica de 36, Obtém-se

a)6^-2/3

b)2^1/3 + 3^1/3

c)3^1/3 - 2^1/3

d)5^1/3

e)6

por Rafa.2604 Ter 12 Jul 2016, 15:44

(2^-2/3 - 3^-2/3) . (raiz cúbica de 3 - raiz cúbica de 2)^-1 . raiz cúbica de 36

$\sqrt[3]{36}. \frac{(2^{-2/3}-3^{-2/3})} {(\sqrt[3]{3}-\sqrt[3]{2})} = 36^{1/3}. \frac{(2^{-2/3}- 3^{-2/3})} {(3^{1/3}- 2^{1/3})} = (2^{2}.3^{2})^{1/3}. \frac{(2^{-2/3}- 3^{-2/3})} {(3^{1/3}- 2^{1/3})} = \\\\ = (2^{2/3}.3^{2/3}). \frac{(2^{-2/3}- 3^{-2/3})} {(3^{1/3}- 2^{1/3})} = \frac{(2^{2/3}.3^{2/3}.2^{-2/3} - 2^{2/3}.3^{2/3}. 3^{-2/3})} {(3^{1/3}- 2^{1/3})} = \frac{(3^{2/3} - 2^{2/3})} {(3^{1/3}- 2^{1/3})} = \\\\ = \frac{(3^{2/3} - 2^{2/3})} {(3^{1/3}- 2^{1/3})} . \frac{(3^{1/3} + 2^{1/3})} {(3^{1/3} + 2^{1/3})} = \frac{(3^{2/3} - 2^{2/3})(3^{1/3} + 2^{1/3})}{(3^{1/3}.3^{1/3}+3^{1/3}.2^{1/3} -2^{1/3}.3^{1/3}-2^{1/3}.2^{1/3})} = \\\\ = \frac{(3^{2/3} - 2^{2/3})(3^{1/3} + 2^{1/3})}{(3^{2/3}+6^{1/3} -6^{1/3} - 2^{2/3})} = \frac{(3^{2/3} - 2^{2/3})(3^{1/3} + 2^{1/3})}{(3^{2/3}- 2^{2/3})} = \;\; 3^{1/3} + 2^{1/3}$

Logo, a alternativa correta é b)2^1/3 + 3^1/3 .