Question

Se n ∈ N e n > 1, então o valor de $\sqrt[n]{ \frac{20}{ 4^{n+2} + 2^{2n+2} } }$ será:
a)$\frac{4}{n}$
b)$\frac{1}{4 \sqrt[n]{2n} }$
c) $\frac{1}{2n}$
d)$\sqrt[n]{2n + 1}$
e) $\frac{1}{4}$

Answer 1

Acho que é assim---
--> 4^(n+2) + 2^(2n+2)=  2^(2n+4)+ 2^(2n+2)
2^(2n) x 2^4  + 2^(2n) x 2²
2^(2n) x (2^4 + 2^2)
2^(2n) x 20 
este 20 corta com o de cima e o (n) no expoente corta com o n do radical.
resposta= 1/4

Answer 2

$\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{(2^{2})^{n+2}+2^{2n+2}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2(n+2)}+2^{2n+2}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$

Colocando 2^(2n + 2) em evidência no denominador:

$\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{(2n+2)}\cdot[2^{2}+1]}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{(2n+2)}\cdot5}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{4}{2^{(2n+2)}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{2^{2}}{2^{(2n+2)}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{2^{2-(2n+2)}}$

$\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{2^{2-2n-2}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{2^{-2n}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\sqrt[n]{\dfrac{1}{2^{2n}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\dfrac{\sqrt[n]{1}}{\sqrt[n]{(2^{2})^{n}}}\\\\\\\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\dfrac{1}{2^{2}}\\\\\\\boxed{\boxed{\sqrt[n]{\dfrac{20}{4^{n+2}+2^{2n+2}}}=\dfrac{1}{4}}}$