Question

Media geometrica a numerelor 12∛3, 288, 108∛9 . Si rezolvarea , va rog.

Answer 1

$m_g=\sqrt[3]{12\sqrt[3]{3}\cdot288\cdot108\cdot\sqrt[3]{3^2}}=\sqrt[3]{2^2\cdot3\cdot3\cdot2^5\cdot3^2\cdot2^2\cdot3^3}=$

$=\sqrt[3]{2^9\cdot3^7}=2^3\cdot3^2\cdot\sqrt[3]3=72\sqrt[3]3$

Am folosit la inceput : $\sqrt[3]3\cdot\sqrt[3]{3^2}=\sqrt[3]{3^3}=3$