Пусть $$A = \sqrt{27 + 10\sqrt{2}} + \sqrt{27 - 10\sqrt{2}}$$.
Тогда $$A^2 = (\sqrt{27 + 10\sqrt{2}} + \sqrt{27 - 10\sqrt{2}})^2 = 27 + 10\sqrt{2} + 2\sqrt{(27 + 10\sqrt{2})(27 - 10\sqrt{2})} + 27 - 10\sqrt{2} = 54 + 2\sqrt{27^2 - (10\sqrt{2})^2} = 54 + 2\sqrt{729 - 200} = 54 + 2\sqrt{529} = 54 + 2 \cdot 23 = 54 + 46 = 100$$.
Так как $$A > 0$$, то $$A = \sqrt{100} = 10$$.
Ответ: 10