Odpowiedź:
[tex]a)\ \ (3\frac{3}{8})^{-\frac{2}{3}}=(\frac{27}{8})^{-\frac{2}{3}}=(\frac{8}{27})^{\frac{2}{3}}=\left(\sqrt[3]{\frac{8}{27}}\right)^2=(\frac{2}{3})^2=\frac{2^2}{3^2}=\frac{4}{9}\\\\b)\ \ \sqrt[4]{\frac{17}{128}}:\sqrt[4]{2\frac{1}{8}}=\sqrt[4]{\frac{17}{128}:2\frac{1}{8}}=\sqrt[4]{\frac{17}{128}:\frac{17}{8}}=\sqrt[4]{\frac{\not17^1}{\not128_{16}}\cdot\frac{\not8^1}{\not17_{1}}}=\sqrt[4]{\frac{1}{16}}=\frac{1}{2}[/tex]
[tex]c)\ \ log_{3}27\sqrt{3}=log_{3}3^3\cdot3^{\frac{1}{2}}=log_{3}3^{3+\frac{1}{2}}=log_{3}3^{\frac{7}{2}}=\frac{7}{2}=3\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
Wykorzystano własności pierwiastków
[tex](\frac{a}{b})^{-n}=(\frac{b}{a})^n[/tex]
[tex]a^{\frac{m}{n}}=(\sqrt[n]{a})^m\\\\\sqrt[n]{a}:\sqrt[n]{b}=\sqrt[n]{a:b}[/tex]
----------------------------
[tex]log_{a}a=1\\\\log_{a}a^x=x[/tex]
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Odpowiedź:
[tex]a)\ \ (3\frac{3}{8})^{-\frac{2}{3}}=(\frac{27}{8})^{-\frac{2}{3}}=(\frac{8}{27})^{\frac{2}{3}}=\left(\sqrt[3]{\frac{8}{27}}\right)^2=(\frac{2}{3})^2=\frac{2^2}{3^2}=\frac{4}{9}\\\\b)\ \ \sqrt[4]{\frac{17}{128}}:\sqrt[4]{2\frac{1}{8}}=\sqrt[4]{\frac{17}{128}:2\frac{1}{8}}=\sqrt[4]{\frac{17}{128}:\frac{17}{8}}=\sqrt[4]{\frac{\not17^1}{\not128_{16}}\cdot\frac{\not8^1}{\not17_{1}}}=\sqrt[4]{\frac{1}{16}}=\frac{1}{2}[/tex]
[tex]c)\ \ log_{3}27\sqrt{3}=log_{3}3^3\cdot3^{\frac{1}{2}}=log_{3}3^{3+\frac{1}{2}}=log_{3}3^{\frac{7}{2}}=\frac{7}{2}=3\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
Wykorzystano własności pierwiastków
[tex](\frac{a}{b})^{-n}=(\frac{b}{a})^n[/tex]
[tex]a^{\frac{m}{n}}=(\sqrt[n]{a})^m\\\\\sqrt[n]{a}:\sqrt[n]{b}=\sqrt[n]{a:b}[/tex]
----------------------------
[tex]log_{a}a=1\\\\log_{a}a^x=x[/tex]