1. Вычислите: 1)3√32+√-27+√1; 2)√8¹²; 3)⁴√0,0081⋅16; 4)⁶√64/√2;

1) Вычислим: $$3\sqrt{32}+\sqrt[3]{-27}+\sqrt[6]{1}=3\sqrt{16\cdot 2}-3+1=3\cdot 4\sqrt{2}-2=12\sqrt{2}-2$$.

2) $$ \sqrt[4]{8^{12}} = (8^{12})^{\frac{1}{4}} = 8^{\frac{12}{4}} = 8^3 = 512 $$.

3) $$ \sqrt[4]{0,0081 \cdot 16} = \sqrt[4]{\frac{81}{10000} \cdot 16} = \sqrt[4]{\frac{3^4}{10^4} \cdot 2^4} = \frac{3}{10} \cdot 2 = \frac{6}{10} = 0,6 $$.

4) $$ \frac{\sqrt[6]{64}}{\sqrt[3]{2}} = \frac{\sqrt[6]{2^6}}{\sqrt[3]{2}} = \frac{2}{\sqrt[3]{2}} = \frac{2 \cdot \sqrt[3]{2^2}}{\sqrt[3]{2} \cdot \sqrt[3]{2^2}} = \frac{2 \sqrt[3]{4}}{2} = \sqrt[3]{4} $$.

Ответ: 1) $$12\sqrt{2}-2$$, 2) $$512$$, 3) $$0,6$$, 4) $$\sqrt[3]{4}$$.