[tex]a) \ 16^{\frac{5}{12}}\cdot16^{\frac{1}{3}} = 16^{\frac{5}{12}+\frac{1}{3}} = 16^{\frac{5}{12}+\frac{4}{12}} = 16^{\frac{9}{12}} = 16^{\frac{3}{4}} = (2^{4})^{\frac{3}{4}} = 2^{4}\cdot\frac{3}{4}}=2^{3} = 8[/tex]
[tex]c) \ (2\frac{2}{9})^{-\frac{2}{3}}\cdot(\frac{2}{15})^{-\frac{2}{3}} = (\frac{20}{9}\cdot\frac{2}{15})^{-\frac{2}{3}} = (\frac{8}{27})^{-\frac{2}{3}}=(\frac{27}{8})^{\frac{2}{3}} = ((\frac{3}{2})^{3})^{\frac{2}{3}} = (\frac{3}{2})^{3\cdot\frac{2}{3}} =\\\\=(\frac{3}{2})^{2} = \frac{9}{4} = 2\frac{1}{4}[/tex]
[tex]e) \ ((\frac{8}{27})^{-\frac{4}{5}})^{\frac{5}{6}}=((\frac{27}{8})^{\frac{5}{4}})^{\frac{5}{6}} = (((\frac{3}{2})^{3})^{\frac{4}{5}})^{\frac{5}{6}}=(\frac{3}{2})^{3\cdot\frac{4}{5}\cdot\frac{5}{6}} = (\frac{3}{2})^{2} = \frac{9}{4} = 2\frac{1}{4}[/tex]
Wykorzystano własności potęg:
[tex]a^{m}\cdot a^{n} = a^{m+n}\\\\a^{m}:a^{n} = a^{m-n}\\\\(a^{m})^{n} = a^{m\cdot n}\\\\a^{-n} = (\frac{1}{a})^{n}, \ \ dla \ a \neq 0[/tex]
