Question

Вычислите, подробно:

Answer 1

$\tt 1) \ {(2^{\frac{1}{\sqrt{2}}})}^\sqrt{8}}=2^{\frac{1}{\sqrt{2}}\cdot\sqrt{8}}=2^{\sqrt{4}}=2^2=4\\\\\\ 2) \ {(2^{\sqrt{27}})}^\sqrt{3}}\cdot 2^{-3}=2^{3\sqrt{3} \cdot\sqrt{3}}\cdot2^{-3}=2^9\cdot2^{-3}=2^{9+(-3)}=2^6=64$