Question

Упростить выражение:
$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

Answer 1

$\frac{2+ \sqrt{3} }{ \sqrt{2}+ \sqrt{2+ \sqrt{3} } } + \frac{2- \sqrt{3} }{ \sqrt{2}- \sqrt{2- \sqrt{3} } }= \\ \\ \frac{2+ \sqrt{3} }{ \sqrt{2}+ \frac{ \sqrt{2} }{ \sqrt{2} } \sqrt{2+ \sqrt{3} } }+ \frac{2- \sqrt{3} }{ \sqrt{2}- \frac{ \sqrt{2} }{ \sqrt{2} } \sqrt{2- \sqrt{3} } }= \\ \\ \frac{(2+ \sqrt{3}) \sqrt{2} }{2+ \sqrt{4+2 \sqrt{3} } } +\frac{(2- \sqrt{3}) \sqrt{2} }{2- \sqrt{4-2 \sqrt{3} } } =$
$\frac{2 \sqrt{2} + \sqrt{6} }{2+ \sqrt{( \sqrt{3}+1)^2 } } + \frac{2 \sqrt{2} - \sqrt{6} }{2- \sqrt{( \sqrt{3}-1)^2 } } =\\ \\ \frac{2 \sqrt{2} + \sqrt{6} }{2+ \sqrt{3} +1 } } + \frac{2 \sqrt{2} - \sqrt{6} }{2- \sqrt{3} +1 } }= \\ \\ \frac{2 \sqrt{2} + \sqrt{6} }{3+ \sqrt{3} } }+ \frac{2 \sqrt{2} - \sqrt{6} }{3- \sqrt{3} } }= \\ \\ \frac{(2 \sqrt{2} + \sqrt{6})(3- \sqrt{3}) +(2 \sqrt{2} - \sqrt{6})(3+ \sqrt{3})}{(3+ \sqrt{3} )(3- \sqrt{3} )} =$
$\frac{6 \sqrt{2} +3 \sqrt{6}-2 \sqrt{6} - \sqrt{18} +6 \sqrt{2} -3 \sqrt{6}+2 \sqrt{6} - \sqrt{18} }{9-3} = \\ \\ \frac{12 \sqrt{2}-6 \sqrt{2} }{6} = \sqrt{2 \\ \\ }$

Answer 2

task/28500440
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ответ : √2