Odpowiedź:
[tex]a)\ \ 4\sqrt{3}\cdot\sqrt{2}=4\sqrt{3\cdot2}=4\sqrt{6}\\\\b)\ \ \frac{\not12^3\sqrt{15}}{\not4_{1}\sqrt{3}}=\frac{3\sqrt{15}}{\sqrt{3}}=3\sqrt{\frac{15}{3}}=3\sqrt{5}\\\\c)\ \ 2\sqrt[3]{6}\cdot\sqrt[3]{2}=2\sqrt[3]{6\cdot2}=2\sqrt[3]{12}\\\\d)\ \ \frac{\not8^4\sqrt[3]{-20}}{\not2_{1}\sqrt[3]{5}}=\frac{4(-\sqrt[3]{20})}{\sqrt[3]{5}}=\frac{-4\sqrt[3]{20}}{\sqrt[3]{5}}=-4\sqrt[3]{\frac{20}{5}}=-4\sqrt[3]{4}[/tex]
