Question

$\sqrt[3]{ \frac{2^{31}+ 2^{33} }{10} }$

Como faço para colocar em evidencia o $2^{31}+ 2^{33}$ para que ele fique $( 2^{30} * (2^{1} + 2^{3} )$?

Answer 1

$E=\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}+2^{33}}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}+2^{31+2}}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}+2^{31}\cdot 2^{2}}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}\cdot 1+2^{31}\cdot 2^{2}}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}\cdot (1+2^{2})}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}\cdot (1+4)}{10}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}\cdot 5}{10}}$

$=\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}\cdot \diagup\!\!\!\! 5}{2\cdot \diagup\!\!\!\! 5}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{\dfrac{2^{31}}{2^{1}}}\\\\\\ =\,^{3}\!\!\!\sqrt{2^{31-1}}\\\\\\ =\,^{3}\!\!\!\sqrt{2^{30}}\\\\\\ =\,^{3}\!\!\!\sqrt{2^{10\,\cdot\,3}}\\\\\\ =\,^{3}\!\!\!\!\sqrt{(2^{10})^3}\\\\\\ =2^{10}\\\\ =\boxed{\begin{array}{c}1\,024 \end{array}}$


Bons estudos! :-)