Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 69

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\[1)\ 2^{2 - 3\sqrt{5}} \bullet 8^{\sqrt{5}} =\]

\[= 2^{2 - 3\sqrt{5}} \bullet \left( 2^{3} \right)^{\sqrt{5}} =\]

\[= 2^{2 - 3\sqrt{5}} \bullet 2^{3\sqrt{5}} = 2^{2 - 3\sqrt{5} + 3\sqrt{5}} =\]

\[= 2^{2} = 4\]

\[2)\ 3^{1 + 2\sqrt[3]{2}}\ :9^{\sqrt[3]{2}} =\]

\[= 3^{1 + 2\sqrt[3]{2}}\ :\left( 3^{2} \right)^{\sqrt[3]{2}} =\]

\[= 3^{1 + 2\sqrt[3]{2}}\ :3^{2\sqrt[3]{2}} = 3^{1 + 2\sqrt[3]{2} - 2\sqrt[3]{2}} =\]

\[= 3^{1} = 3\]

\[3)\ \left( 5^{1 + \sqrt{2}} \right)^{1 - \sqrt{2}} =\]

\[= 5^{\left( 1 + \sqrt{2} \right)\left( 1 - \sqrt{2} \right)} = 5^{1^{2} - \left( \sqrt{2} \right)^{2}} =\]

\[= 5^{1 - 2} = 5^{- 1} = \frac{1}{5} = 0,2\]

\[4)\ \left( 5^{1 - \sqrt{5}} \right)^{1 + \sqrt{5}} - \left( \sqrt{5} \right)^{0} =\]

\[= 5^{\left( 1 - \sqrt{5} \right)\left( 1 + \sqrt{5} \right)} - 1 =\]

\[= 5^{1^{2} - \left( \sqrt{5} \right)^{2}} - 1 = 5^{1 - 5} - 1 =\]

\[= 5^{- 4} - 1 = \frac{1}{5^{4}} - 1 =\]

\[= \frac{1}{625} - \frac{625}{625} = - \frac{624}{625}\]