A expressão √8.(∛(√2) )^6/(√2) é igual a?
A)4√2 B)2√2 C)8 D)4 E)NDA
Anexos:
adrielcavalcant:
Oi
Oi
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![\frac{ \sqrt[2]{8} * (\sqrt[3]{ \sqrt[2]{2}} )^{6}}{ \sqrt{2} } = \\ \\ \frac{8^{ \frac{1}{2} } * (\sqrt[2]{2} )^{ \frac{6}{3} }}{\sqrt{2}} = \\ \\ \frac{ 2^{(3* \frac{1}{2}) } *(\sqrt[2]{2} )^{2}}{ 2^{ \frac{1}{2} } } = \\ \\ \frac{2^{ \frac{3}{2}} * 2^{ \frac{2}{2} }}{2^{ \frac{1}{2}} } = \frac{2^{ \frac{5}{2} }}{2^{ \frac{1}{2}} } = \\ \\ 2^{ \frac{5}{2} - \frac{1}{2} } = 2^{ \frac{4}{2} } = 2 ^{2} = 4](https://tex.z-dn.net/?f=%5Cfrac%7B+%5Csqrt%5B2%5D%7B8%7D+%2A+%28%5Csqrt%5B3%5D%7B+%5Csqrt%5B2%5D%7B2%7D%7D+%29%5E%7B6%7D%7D%7B+%5Csqrt%7B2%7D+%7D++%3D+%5C%5C+%5C%5C++%5Cfrac%7B8%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%2A+%28%5Csqrt%5B2%5D%7B2%7D+%29%5E%7B+%5Cfrac%7B6%7D%7B3%7D+%7D%7D%7B%5Csqrt%7B2%7D%7D++%3D+%5C%5C+%5C%5C+%5Cfrac%7B+2%5E%7B%283%2A+%5Cfrac%7B1%7D%7B2%7D%29+%7D+%2A%28%5Csqrt%5B2%5D%7B2%7D+%29%5E%7B2%7D%7D%7B+2%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D+%7D++%3D+%5C%5C+%5C%5C+%5Cfrac%7B2%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%2A+2%5E%7B+%5Cfrac%7B2%7D%7B2%7D+%7D%7D%7B2%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D+%7D++%3D++%5Cfrac%7B2%5E%7B+%5Cfrac%7B5%7D%7B2%7D+%7D%7D%7B2%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D+%7D+%3D++%5C%5C+%5C%5C++2%5E%7B+%5Cfrac%7B5%7D%7B2%7D+-++%5Cfrac%7B1%7D%7B2%7D+%7D+++%3D++2%5E%7B+%5Cfrac%7B4%7D%7B2%7D+%7D+%3D+2+%5E%7B2%7D+++%3D+4 "\frac{ \sqrt[2]{8} * (\sqrt[3]{ \sqrt[2]{2}} )^{6}}{ \sqrt{2} } = \\ \\ \frac{8^{ \frac{1}{2} } * (\sqrt[2]{2} )^{ \frac{6}{3} }}{\sqrt{2}} = \\ \\ \frac{ 2^{(3* \frac{1}{2}) } *(\sqrt[2]{2} )^{2}}{ 2^{ \frac{1}{2} } } = \\ \\ \frac{2^{ \frac{3}{2}} * 2^{ \frac{2}{2} }}{2^{ \frac{1}{2}} } = \frac{2^{ \frac{5}{2} }}{2^{ \frac{1}{2}} } = \\ \\ 2^{ \frac{5}{2} - \frac{1}{2} } = 2^{ \frac{4}{2} } = 2 ^{2} = 4")
Até mais !
Até mais !
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