Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 61

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\[1)\ \sqrt[3]{a} \bullet \sqrt[6]{a} = a^{\frac{1}{3}} \bullet a^{\frac{1}{6}} = a^{\frac{1}{3} + \frac{1}{6}} =\]

\[= a^{\frac{2 + 1}{6}} = a^{\frac{3}{6}} = a^{\frac{1}{2}} = \sqrt{a}\]

\[Если\ a = 0,09:\]

\[\sqrt{a} = \sqrt{0,09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0,3.\]

\[2)\ \sqrt{b}\ :\sqrt[6]{b} = b^{\frac{1}{2}}\ :b^{\frac{1}{6}} = b^{\frac{1}{2} - \frac{1}{6}} =\]

\[= b^{\frac{3 - 1}{6}} = b^{\frac{2}{6}} = b^{\frac{1}{3}} = \sqrt[3]{b};\]

\[Если\ b = 27:\]

\[\sqrt[3]{b} = \sqrt[3]{27} = \sqrt[3]{3^{3}} = 3.\]

\[3)\ \frac{\sqrt{b} \bullet \sqrt[3]{b^{2}}}{\sqrt[6]{b}} = \frac{b^{\frac{1}{2}} \bullet b^{\frac{2}{3}}}{b^{\frac{1}{6}}} =\]

\[= b^{\frac{1}{2} + \frac{2}{3} - \frac{1}{6}} = b^{\frac{3 + 4 - 1}{6}} = b^{\frac{6}{6}} = b\]

\[b = 1,3.\]

\[4)\ \sqrt[3]{a} \bullet \sqrt[4]{a} \bullet \sqrt[12]{a^{5}} =\]

\[= a^{\frac{1}{3}} \bullet a^{\frac{1}{4}} \bullet a^{\frac{5}{12}} = a^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}} =\]

\[= a^{\frac{4 + 3 + 5}{12}} = a^{\frac{12}{12}} = a;\]

\[\ a = 2,7.\]