Question

покажите, как делать, п16???

Answer 1

$\sqrt{2} \cdot (\sqrt{8 + \sqrt{2} \cdot \sqrt{9+ \sqrt{17}}}}+ \sqrt{8 - \sqrt{2} \cdot \sqrt{9- \sqrt{17}}})= \\\\ =\sqrt{2} \cdot ( \sqrt{8 + \sqrt{18+ 2\sqrt{17}}}}+ \sqrt{8 - \sqrt{18- 2\sqrt{17}}})= \\\\ =\sqrt{2} \cdot (\sqrt{8 + \sqrt{1+ 2 \cdot 1 \cdot \sqrt{17} + 17}}}+ \sqrt{8 -\sqrt{1- 2 \cdot 1 \sqrt{17} + 17}})= \\ \\ =\sqrt{2} \cdot (\sqrt{8 + \sqrt{(1+\sqrt{17})^2}}}+ \sqrt{8 - \sqrt{(1- \sqrt{17})^2}})=$

$\\ \\ =\sqrt{2} \cdot (\sqrt{8 +1+\sqrt{17}}}}+ \sqrt{8 -( \sqrt{17}-1)}}) =\\ \\ =\sqrt{2} \cdot (\sqrt{9+\sqrt{17}}}}+ \sqrt{9- \sqrt{17}}})=\sqrt{18+2\sqrt{17}}}}+ \sqrt{18- 2\sqrt{17}}}=\\\\= \sqrt{1+2\sqrt{17}+17}}}+ \sqrt{1- 2\sqrt{17}+ 17}}= \sqrt{(1+ \sqrt{17})^2}+ \sqrt{(1 - \sqrt{17})^2} \\\\ = 1+\sqrt{17} + \sqrt{17} -1 =2\sqrt{17}$