Question

Срочно, алгебра
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Answer 1

$\displaystyle\tt \frac{1}{2+\sqrt{6}}+ \frac{1}{2+\sqrt{6}}=\frac{1(2+\sqrt{6})+1(2-\sqrt{6})}{(2+\sqrt{6})(2-\sqrt{6})}=\frac{2+\sqrt{6}+2-\sqrt{6}}{2^2-(\sqrt{6})^2}=\\\\\\ =\frac{4}{4-6} =\frac{4}{-2} =-2$