Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 12

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\[1)\ \sqrt{\left( \sqrt{7 - 2\sqrt{10}} + \sqrt{2} \right) \bullet 2\sqrt{5}} =\]

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

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

\[= \sqrt{\sqrt{5} \bullet 2\sqrt{5}} = \sqrt{2\sqrt{25}} =\]

\[= \sqrt{2 \bullet 5} = \sqrt{10}\]

\[2)\ \sqrt{\left( \sqrt{16 - 6\sqrt{7}} + \sqrt{7} \right) \bullet 3} =\]

\[= \sqrt{\left( \sqrt{\left( 3 - \sqrt{7} \right)^{2}} + \sqrt{7} \right) \bullet 3} =\]

\[= \sqrt{\left( 3 - \sqrt{7} + \sqrt{7} \right) \bullet 3} =\]

\[= \sqrt{3 \bullet 3} = 3\]

\[= \sqrt{2\sqrt{3} \bullet 2 + 7} = \sqrt{4\sqrt{3} + 7} =\]

\[= \sqrt{4 + 2 \bullet 2\sqrt{3} + 3} =\]

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