Question

√√$\sqrt \sqrt\frac{2^{7}*2^{-1} }{2^{2} }$

Answer 1

Explicação passo a passo:

$\sqrt{\sqrt{\sqrt{\sqrt{{2^7.2^{-1}\over2^2}} } } } =\\ \\\\ \sqrt[16]{{2^{7-1}\over2^2}} =\\ \\ \\ \sqrt[16]{2^6\over2^2}} =\\ \\ \\ \sqrt[16]{2^{6-2}} =\\ \\ \sqrt[16]{2^4} =\\ \\ simplificando\\ \\ \sqrt[16:4]{2^{4:4}} =\boxed{\sqrt[4]{2} }$