Решение выражений:

a) (√2+√3)⋅(√2-√3) = (√2)^2 - (√3)^2 = 2 - 3 = -1

б) (√2+2√3)⋅(√2-√3) = (√2)^2 - √2√3 + 2√3√2 - (2√3)^2 = 2 + √6 - 4*3 = 2 + √6 - 12 = -10 + √6

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

г) $$\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{1}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{2\sqrt{2}}{3 - 2} = 2\sqrt{2} =$$ 2√2

д) $$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} = \frac{(\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{(\sqrt{2} + \sqrt{3})^2}{3 - 2} = (\sqrt{2})^2 + 2\sqrt{2}\sqrt{3} + (\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 =$$ 5 + 2√6

е) $$\frac{\sqrt{5}}{\sqrt{5} - \sqrt{2}} + \frac{\sqrt{5}}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5}(\sqrt{5} + \sqrt{2}) + \sqrt{5}(\sqrt{5} - \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{5 + \sqrt{10} + 5 - \sqrt{10}}{5 - 2} = \frac{10}{3} =$$ 10/3