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:

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

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