Question

$\sqrt[n]{\frac{36 (3^{8n)} }{27^{2n+1} + 9^{3n+1} } }$

Answer 1

Respuesta:

9

Explicación paso a paso:

$\sqrt[n]{ \frac{ 36({3}^{8n}) }{ {27}^{2n + 1} + {9}^{3n + 1} } } \\ \sqrt[n]{ \frac{4. {3}^{2} ( {3}^{8n} )}{ {( {3}^{3} )}^{2n + 1} + { ({3}^{2}) }^{3n + 1} } } \\ \sqrt[n]{ \frac{4. {3}^{8n + 2} }{ {3}^{6n + 3} + {3}^{6n + 2} } } \\ \sqrt[n]{ \frac{ {3}^{6n + 2} . {3}^{2n} .4}{ {3}^{6n + 2} ( {3}^{1} + 1) } } \\simplificando \\ \sqrt[n]{ \frac{ {3}^{2n} .4}{(3 + 1)} } \\ \sqrt[n]{ \frac{ {3}^{2n} .4}{4} } \\ simplificando \\ \sqrt[n]{ {3}^{2n} } = {3}^{ \frac{2n}{n} } = {3}^{2} = 9$