Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 71

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\[1)\ \frac{10^{2 + \sqrt{7}}}{2^{2 + \sqrt{7}} \bullet 5^{1 + \sqrt{7}}} =\]

\[= \frac{(2 \bullet 5)^{2 + \sqrt{7}}}{2^{2 + \sqrt{7}} \bullet 5^{1 + \sqrt{7}}} = \frac{2^{2 + \sqrt{7}} \bullet 5^{2 + \sqrt{7}}}{2^{2 + \sqrt{7}} \bullet 5^{1 + \sqrt{7}}} =\]

\[= 2^{2 + \sqrt{7} - \left( 2 + \sqrt{7} \right)} \bullet 5^{2 + \sqrt{7} - \left( 1 + \sqrt{7} \right)} =\]

\[= 2^{0} \bullet 5^{1} = 1 \bullet 5 = 5\]

\[2)\ \frac{6^{3 + \sqrt{5}}}{2^{2 + \sqrt{5}} \bullet 3^{1 + \sqrt{5}}} =\]

\[= \frac{(2 \bullet 3)^{3 + \sqrt{5}}}{2^{2 + \sqrt{5}} \bullet 3^{1 + \sqrt{5}}} = \frac{2^{3 + \sqrt{5}} \bullet 3^{3 + \sqrt{5}}}{2^{2 + \sqrt{5}} \bullet 3^{1 + \sqrt{5}}} =\]

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

\[= 2^{1} \bullet 3^{2} = 2 \bullet 9 = 18\]

\[3)\ \left( 25^{1 + \sqrt{2}} - 5^{2\sqrt{2}} \right) \bullet 5^{- 1 - 2\sqrt{2}} =\]

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

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

\[= 5^{1} - 5^{- 1} = 5 - \frac{1}{5} = \frac{25}{5} - \frac{1}{5} =\]

\[= \frac{24}{5} = 4\frac{4}{5} = 4,8\]

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

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

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

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

\[= 2^{0} - 2^{- 2} = 1 - \frac{1}{2^{2}} = 1 - \frac{1}{4} =\]

\[= \frac{4}{4} - \frac{1}{4} = \frac{3}{4} = 0,75\]