Question

$\sqrt[n]{ \frac{ {3}^ {n + 1 } }{ \sqrt[n + 2]{9} \sqrt{ {9}^{n} } } }$
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Answer 1

Hola!

$\\ \sqrt[n]{ \frac{ {3}^ {n + 1 } }{ \sqrt[n + 2]{9} \sqrt{ {9}^{n} } } } \\ \sqrt[n]{ \frac{ {3}^ {n + 1 } }{ \sqrt[n + 2]{ {3}^{2} } \sqrt{ {3}^{2n} } } } \:$

$\sqrt[n]{ \frac{ {3}^ {n + 1 } }{ {3}^{ \frac{2}{n + 2} } {3}^{ \frac{2n}{2} } } } \\ \sqrt[n]{ \frac{ {3}^ {n + 1 } }{ {3}^{ \frac{2}{n + 2} } {3}^{n} } }$

$\sqrt[n]{ \frac{ {3}^ {n + 1 } }{ {3}^{ \frac{2}{n + 2} + n} }} \\ \sqrt[n]{ \frac{ {3}^{n + 1} }{ {3}^{ \frac{ {n}^{2} + 2n + 2 }{n + 2} } } }$

$\\ \sqrt[n]{ {3}^{(n + 1) - (\frac{ {n}^{2} + 2n + 2 }{n + 2}) } } \\ \sqrt[n]{ {3}^{ \frac{( {n}^{2} + 3n + 2) - ( {n}^{2} + 2n + 2}{n + 2} } } \\ \sqrt[n]{ {3}^{ \frac{n}{n + 2} } }$

$\\ {3}^{ \frac{ \frac{n}{n + 2} }{n} } \\ {3}^{ \frac{1}{n + 2} } = \sqrt[n + 2]{3}$