Докажите, что:
$\sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{2 + \sqrt{2 + \sqrt{2}}} * \sqrt{2 - \sqrt{2 + \sqrt{2}}} = 2$.
$\sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{2 + \sqrt{2 + \sqrt{2}}} * \sqrt{2 - \sqrt{2 + \sqrt{2}}} = \sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{(2 + \sqrt{2 + \sqrt{2}})(2 - \sqrt{2 + \sqrt{2}})} = \sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{2^2 - (\sqrt{2 + \sqrt{2}})^2} = \sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{4 - (2 + \sqrt{2})} = \sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{4 - 2 - \sqrt{2}} = \sqrt{2} * \sqrt{2 + \sqrt{2}} * \sqrt{2 - \sqrt{2}} = \sqrt{2} * \sqrt{(2 + \sqrt{2})(2 - \sqrt{2})} = \sqrt{2} * \sqrt{2^2 - (\sqrt{2})^2} = \sqrt{2} * \sqrt{4 - 2} = \sqrt{2} * \sqrt{2} = (\sqrt{2})^2 = 2$
Пожауйста, оцените решение
