1.219. Найдите значение выражения: a) 9/(√13-2) + 3/(√13+2); б) 42/(2√6-√3) + 24/(2√3+√6); 8/(√7-√5) - 10/(5-2√5); г) √6+√2 - 9/(√6-√3).

Решение:

a) \( \frac{9}{\sqrt{13} - 2} + \frac{3}{\sqrt{13} + 2} = \frac{9(\sqrt{13} + 2) + 3(\sqrt{13} - 2)}{(\sqrt{13} - 2)(\sqrt{13} + 2)} = \frac{9\sqrt{13} + 18 + 3\sqrt{13} - 6}{13 - 4} = \frac{12\sqrt{13} + 12}{9} = \frac{4\sqrt{13} + 4}{3} \)

б) \( \frac{42}{2\sqrt{6} - \sqrt{3}} + \frac{24}{2\sqrt{3} + \sqrt{6}} = \frac{42(2\sqrt{6} + \sqrt{3})}{(2\sqrt{6} - \sqrt{3})(2\sqrt{6} + \sqrt{3})} + \frac{24(2\sqrt{3} - \sqrt{6})}{(2\sqrt{3} + \sqrt{6})(2\sqrt{3} - \sqrt{6})} = \frac{42(2\sqrt{6} + \sqrt{3})}{24 - 3} + \frac{24(2\sqrt{3} - \sqrt{6})}{12 - 6} = \frac{42(2\sqrt{6} + \sqrt{3})}{21} + \frac{24(2\sqrt{3} - \sqrt{6})}{6} = 2(2\sqrt{6} + \sqrt{3}) + 4(2\sqrt{3} - \sqrt{6}) = 4\sqrt{6} + 2\sqrt{3} + 8\sqrt{3} - 4\sqrt{6} = 10\sqrt{3} \)

в) \( \frac{8}{\sqrt{7} - \sqrt{5}} - \frac{10}{5 - 2\sqrt{5}} = \frac{8(\sqrt{7} + \sqrt{5})}{7 - 5} - \frac{10(5 + 2\sqrt{5})}{(5 - 2\sqrt{5})(5 + 2\sqrt{5})} = \frac{8(\sqrt{7} + \sqrt{5})}{2} - \frac{10(5 + 2\sqrt{5})}{25 - 20} = 4(\sqrt{7} + \sqrt{5}) - \frac{10(5 + 2\sqrt{5})}{5} = 4\sqrt{7} + 4\sqrt{5} - 2(5 + 2\sqrt{5}) = 4\sqrt{7} + 4\sqrt{5} - 10 - 4\sqrt{5} = 4\sqrt{7} - 10 \)

г) \( \sqrt{6} + \sqrt{2} - \frac{9}{\sqrt{6} - \sqrt{3}} = \sqrt{6} + \sqrt{2} - \frac{9(\sqrt{6} + \sqrt{3})}{6 - 3} = \sqrt{6} + \sqrt{2} - \frac{9(\sqrt{6} + \sqrt{3})}{3} = \sqrt{6} + \sqrt{2} - 3(\sqrt{6} + \sqrt{3}) = \sqrt{6} + \sqrt{2} - 3\sqrt{6} - 3\sqrt{3} = -2\sqrt{6} + \sqrt{2} - 3\sqrt{3} \)

Ответ: a) (4√13 + 4)/3; б) 10√3; в) 4√7 - 10; г) -2√6 + √2 - 3√3.