Question

Упростить:
$1)\frac{3^{n + 1} * 9^{2n + 2} * 2^{n - 3}}{2^{n + 1} * 81^{n}} \\2)\frac{\sqrt{\sqrt[3]{6}} }{\sqrt{24}} * \sqrt[3]{\sqrt{36}}$

Answer 1

Ответ:

Объяснение:

$1)\ \dfrac{3^{n+1} \cdot 9^{2n+2} \cdot 2^{n-3}}{2^{n+1} \cdot 81^{n}} = \dfrac{3^{n+1} \cdot 3^{4n+4} \cdot 2^{n-3}}{2^{n+1} \cdot 3^{4n}} = \dfrac{3^{5n+5} \cdot 2^{n-3}}{2^{n+1} \cdot 3^{4n}} = \\ \\ = 3^{5n+5-4n} \cdot 2^{n-3-n-1} = 3^{n+5} \cdot 2^{-4} = \frac{1}{16} \cdot 3^{n+5} .$

$2)\ \dfrac{\sqrt{\sqrt[3]{6} }}{\sqrt{24} } \cdot \sqrt[3]{\sqrt{36} } = \dfrac{\sqrt[6]{6} }{\sqrt{24} } \cdot \sqrt[6]{36} } = \dfrac{\sqrt[6]{6^3} }{\sqrt{24} } = \dfrac{\sqrt{6} }{\sqrt{24} } = \dfrac{1}{\sqrt{4} } =0,5$