Odpowiedź:
[tex]\huge\boxed{a) \ \frac{18^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot9^{\frac{1}{5}}}=\frac{1}{6}}[/tex]
Korzystamy z praw potęgowania:
[tex]a^{0} = 1\\\\a^{1} = a\\\\a^{m}\cdot a^{n} =a^{m\cdot n}\\\\a^{m}:a^{n} = a^{m-n}\\\\(a^{m})^{n} = a^{m\cdot n}\\\\a^{n}\cdot b^{n} = (a\cdot b)^{n}\\\\a^{-n} = \frac{1}{n}[/tex]
[tex]a) \ \frac{18^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot9^{\frac{1}{5}}}=\frac{(6\cdot3)^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot(3^{2})^{\frac{1}{5}}} = \frac{6^{\frac{2}{5}}\cdot3^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot3^{\frac{2}{5}}}=6^{\frac{2}{5}-\frac{7}{5}}\cdot3^{\frac{2}{5}-\frac{2}{5}} = 6^{-\frac{5}{5}}\cdot3^{0} =6^{-1}\cdot1 =\boxed{ \frac{1}{6}}[/tex]
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Verified answer
Odpowiedź:
[tex]\huge\boxed{a) \ \frac{18^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot9^{\frac{1}{5}}}=\frac{1}{6}}[/tex]
Potęga
Korzystamy z praw potęgowania:
[tex]a^{0} = 1\\\\a^{1} = a\\\\a^{m}\cdot a^{n} =a^{m\cdot n}\\\\a^{m}:a^{n} = a^{m-n}\\\\(a^{m})^{n} = a^{m\cdot n}\\\\a^{n}\cdot b^{n} = (a\cdot b)^{n}\\\\a^{-n} = \frac{1}{n}[/tex]
[tex]a) \ \frac{18^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot9^{\frac{1}{5}}}=\frac{(6\cdot3)^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot(3^{2})^{\frac{1}{5}}} = \frac{6^{\frac{2}{5}}\cdot3^{\frac{2}{5}}}{6^{\frac{7}{5}}\cdot3^{\frac{2}{5}}}=6^{\frac{2}{5}-\frac{7}{5}}\cdot3^{\frac{2}{5}-\frac{2}{5}} = 6^{-\frac{5}{5}}\cdot3^{0} =6^{-1}\cdot1 =\boxed{ \frac{1}{6}}[/tex]
odp w załączniku ;)