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Radical de ordinu3 din 216+log in baza 3 din 1pe 27 helpp

Răspuns :

Salut,

[tex]\sqrt[3]{216}+log_3\left(\dfrac{1}{27}\right)=\sqrt[3]{6^3}+log_3\left(\dfrac{1}{3^3}\right)=6+log_33^{-3}=\\\\=6-3\cdot log_33=6-3=3.[/tex]

A fost greu ?

Green eyes.