Тема: показательная функция. решите 5 и 7 номера!

Решите задачу:

$(\sqrt{3\frac{3}{5}}-\sqrt{6\frac{2}{5}}):\sqrt{\frac{2}{45}}=(\frac{\sqrt{18}}{\sqrt{5}}-\frac{\sqrt{32}}{\sqrt{5}})*\frac{\sqrt{45}}{\sqrt{2}}=\frac{3\sqrt{2}-4\sqrt{2}}{\sqrt{5}}*\frac{3\sqrt{5}}{\sqrt{2}}= \\ -\frac{\sqrt{2}}{\sqrt{5}}*\frac{3\sqrt{5}}{\sqrt{2}}= -3$

$\sqrt[3]{4-\sqrt5}*\sqrt[3]{4+\sqrt5}*\sqrt[3]{121}=\sqrt[3]{(4-\sqrt5)(4+\sqrt5)*121}= \\ =\sqrt[3]{(4^2-(\sqrt5)^2)*121}=\sqrt[3]{(16-5)*11^2}=\sqrt[3]{11*11^2}=\sqrt[3]{11^3}=11$