Korzystamy z praw potęgowania:
[tex](a^{m})^{n} = a^{m\cdot n}\\\\a^{-n} = \frac{1}{a^{n}}[/tex]
Rozwiązanie
[tex]a) \ 32^{\frac{1}{5}} = (2^{5})^{\frac{1}{5}} = 2^{5\cdot\frac{1}{5}} = 2^{1} = 2\\\\b) \ 64^{\frac{1}{3}} = (4^{3})^{\frac{1}{3}} = 4\\\\c) \ 81^{\frac{1}{2}} = (9^{2})^{\frac{1}{2}} = 9\\\\d) \ 8^{-\frac{1}{3}} = (2^{3})^{-\frac{1}{3}} = 2^{-1} = \frac{1}{2}\\\\e) \ 25^{-\frac{1}{2}} = (5^{2})^{-\frac{1}{2}} = 5^{-1} = \frac{1}{5}\\\\f) \ 243^{-\frac{1}{5}} = (3^{5})^{-\frac{1}{5}} = 3^{-1} = \frac{1}{3}[/tex]
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Korzystamy z praw potęgowania:
[tex](a^{m})^{n} = a^{m\cdot n}\\\\a^{-n} = \frac{1}{a^{n}}[/tex]
Rozwiązanie
[tex]a) \ 32^{\frac{1}{5}} = (2^{5})^{\frac{1}{5}} = 2^{5\cdot\frac{1}{5}} = 2^{1} = 2\\\\b) \ 64^{\frac{1}{3}} = (4^{3})^{\frac{1}{3}} = 4\\\\c) \ 81^{\frac{1}{2}} = (9^{2})^{\frac{1}{2}} = 9\\\\d) \ 8^{-\frac{1}{3}} = (2^{3})^{-\frac{1}{3}} = 2^{-1} = \frac{1}{2}\\\\e) \ 25^{-\frac{1}{2}} = (5^{2})^{-\frac{1}{2}} = 5^{-1} = \frac{1}{5}\\\\f) \ 243^{-\frac{1}{5}} = (3^{5})^{-\frac{1}{5}} = 3^{-1} = \frac{1}{3}[/tex]