ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 70

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\[1)\ 2^{1 - 2\sqrt{2}} \bullet 4^{\sqrt{2}} =\]

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

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

\[= 2^{1} = 2\]

\[2)\ 3^{2 - 3\sqrt{3}} \bullet 27^{\sqrt{3}} =\]

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

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

\[= 3^{2} = 9\]

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

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

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

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

\[4)\ 4^{3 + \sqrt{2}} \bullet 2^{1 - \sqrt{2}} \bullet 2^{- 4 - \sqrt{2}} =\]

\[= \left( 2^{2} \right)^{3 + \sqrt{2}} \bullet 2^{1 - \sqrt{2} + \left( - 4 - \sqrt{2} \right)} =\]

\[= 2^{6 + 2\sqrt{2}} \bullet 2^{- 3 - 2\sqrt{2}} =\]

\[= 2^{6 + 2\sqrt{2} + \left( - 3 - 2\sqrt{2} \right)} = 2^{3} = 8\]