Răspuns:
Explicație pas cu pas:
[tex]\it \sqrt{\dfrac{36}{49}}=\dfrac{6}{7}\\ \\ \\ \sqrt[3]{-\dfrac{\it216}{\it125}}=\sqrt[3]{-\dfrac{\it6^3}{\it5^3}}=-\dfrac{6}{5}\\ \\ \\ \sqrt[3]{\it512^{-1}}=\sqrt[3]{(\it8^3)^{-1}}=8^{-1}=\dfrac{1}{8}\\ \\ \\ \sqrt[3]{\it-0,008}=-0,2[/tex]
