Szczegółowe wyjaśnienie:
Korzystamy z praw potęgowania:
[tex](a^{m})^{n} = a^{m\cdot n}\\\\a^{m}\cdot a^{n}=a^{m+n}[/tex]
[tex]d) \ 3\sqrt[3]{7}\cdot2(\sqrt[3]{7})^{2} = 3\cdot2\cdot7^{\frac{1}{3}}\cdot(7^{\frac{1}{3}})^{2}=6\cdot7^{\frac{1}{3}}\cdot7^{\frac{1}{3}\cdot2} = 6\cdot7^{\frac{1}{3}}\cdot7^{\frac{2}{3}}=6\cdot7^{\frac{1}{3}+\frac{2}{3}}=6\cdot7^{\frac{3}{3}} = \\\\=6\cdot7^{1} = 6\cdot7 = \boxed{42}[/tex]
[tex]e) \ 5(\sqrt[3]{-6})^{2}\cdot4\sqrt[3]{-6} = 5\cdot4\cdot((-6)^{\frac{1}{3}})^{2}\cdot(-6)^{\frac{1}{3}} = 20\cdot(-6)^{\frac{2}{3}}\cdot(-6)^{\frac{1}{3}} = 20\cdot(-6)^{\frac{2}{3}+\frac{1}{3}} =\\\\=20\cdot(-6)^{1} = 20\cdot(-6) =\boxed{ -120}[/tex]
[tex]f) \ (-7\sqrt[3]{-2})^{2}\cdot\sqrt[3]{-2}=(-7)^{2}\cdot((-2)^{\frac{1}{3}})^{2}\cdot(-2)^{\frac{1}{3}} = 49\cdot(-2)^{\frac{2}{3}}\cdot(-2)^{\frac{1}{3}} =\\\\= 49\cdot(-2)^{\frac{2}{3}+\frac{1}{3}}=49\cdot(-2)^{1} = 49\cdot(-2) =\boxed{ -98}[/tex]
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Szczegółowe wyjaśnienie:
Potęgi
Korzystamy z praw potęgowania:
[tex](a^{m})^{n} = a^{m\cdot n}\\\\a^{m}\cdot a^{n}=a^{m+n}[/tex]
[tex]d) \ 3\sqrt[3]{7}\cdot2(\sqrt[3]{7})^{2} = 3\cdot2\cdot7^{\frac{1}{3}}\cdot(7^{\frac{1}{3}})^{2}=6\cdot7^{\frac{1}{3}}\cdot7^{\frac{1}{3}\cdot2} = 6\cdot7^{\frac{1}{3}}\cdot7^{\frac{2}{3}}=6\cdot7^{\frac{1}{3}+\frac{2}{3}}=6\cdot7^{\frac{3}{3}} = \\\\=6\cdot7^{1} = 6\cdot7 = \boxed{42}[/tex]
[tex]e) \ 5(\sqrt[3]{-6})^{2}\cdot4\sqrt[3]{-6} = 5\cdot4\cdot((-6)^{\frac{1}{3}})^{2}\cdot(-6)^{\frac{1}{3}} = 20\cdot(-6)^{\frac{2}{3}}\cdot(-6)^{\frac{1}{3}} = 20\cdot(-6)^{\frac{2}{3}+\frac{1}{3}} =\\\\=20\cdot(-6)^{1} = 20\cdot(-6) =\boxed{ -120}[/tex]
[tex]f) \ (-7\sqrt[3]{-2})^{2}\cdot\sqrt[3]{-2}=(-7)^{2}\cdot((-2)^{\frac{1}{3}})^{2}\cdot(-2)^{\frac{1}{3}} = 49\cdot(-2)^{\frac{2}{3}}\cdot(-2)^{\frac{1}{3}} =\\\\= 49\cdot(-2)^{\frac{2}{3}+\frac{1}{3}}=49\cdot(-2)^{1} = 49\cdot(-2) =\boxed{ -98}[/tex]