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Решите задачу:

$\sqrt{2+ \sqrt{3} } \sqrt{2+ \sqrt{2+ \sqrt{3} } } \sqrt{2- \sqrt{2+ \sqrt{3} } } = \\\ = \sqrt{2+ \sqrt{3} } \sqrt{(2+ \sqrt{2+ \sqrt{3} })(2- \sqrt{2+ \sqrt{3} } )} = \\\ = \sqrt{2+ \sqrt{3} } \sqrt{4- 2- \sqrt{3}} =\sqrt{2+ \sqrt{3} } \sqrt{ 2- \sqrt{3}} = \\\ =\sqrt{(2+ \sqrt{3})( 2- \sqrt{3})} =\sqrt{4- 3} = \sqrt{1} =1$

$( \sqrt{11+6 \sqrt{2} }+ \sqrt{11-6 \sqrt{2} })^2= \\\ =11+6 \sqrt{2} + 11-6 \sqrt{2} +2 \sqrt{(11+6 \sqrt{2})(11-6 \sqrt{2})} = \\\ =22 +2 \sqrt{121-72} =22 +2 \sqrt{49} =22 +14=36$