1.219. Найдите значение выражения: a) 9/(√13-2) + 3/(4+√13); б) (8/(√7-√5)) - (10/(5-2√5)); в) 42/(2√6-√3) + 24/(2√3+√6); г) 8/(√6+√2) - 9/(√6-√3).

Решение:

a) \( \frac{9}{\sqrt{13}-2} + \frac{3}{4+\sqrt{13}} \)

Первая дробь: \( \frac{9}{\sqrt{13}-2} = \frac{9(\sqrt{13}+2)}{(\sqrt{13}-2)(\sqrt{13}+2)} = \frac{9(\sqrt{13}+2)}{13-4} = \frac{9(\sqrt{13}+2)}{9} = \sqrt{13}+2 \)

Вторая дробь: \( \frac{3}{4+\sqrt{13}} = \frac{3(4-\sqrt{13})}{(4+\sqrt{13})(4-\sqrt{13})} = \frac{3(4-\sqrt{13})}{16-13} = \frac{3(4-\sqrt{13})}{3} = 4-\sqrt{13} \)

Сумма: \( (\sqrt{13}+2) + (4-\sqrt{13}) = \sqrt{13}+2+4-\sqrt{13} = 6 \)

б) \( \frac{8}{\sqrt{7}-\sqrt{5}} - \frac{10}{5-2\sqrt{5}} \)

Первая дробь: \( \frac{8}{\sqrt{7}-\sqrt{5}} = \frac{8(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})} = \frac{8(\sqrt{7}+\sqrt{5})}{7-5} = \frac{8(\sqrt{7}+\sqrt{5})}{2} = 4(\sqrt{7}+\sqrt{5}) = 4\sqrt{7}+4\sqrt{5} \)

Вторая дробь: \( \frac{10}{5-2\sqrt{5}} = \frac{10(5+2\sqrt{5})}{(5-2\sqrt{5})(5+2\sqrt{5})} = \frac{10(5+2\sqrt{5})}{5^2 - (2\sqrt{5})^2} = \frac{10(5+2\sqrt{5})}{25 - 4 \cdot 5} = \frac{10(5+2\sqrt{5})}{25 - 20} = \frac{10(5+2\sqrt{5})}{5} = 2(5+2\sqrt{5}) = 10+4\sqrt{5} \)

Разность: \( (4\sqrt{7}+4\sqrt{5}) - (10+4\sqrt{5}) = 4\sqrt{7}+4\sqrt{5}-10-4\sqrt{5} = 4\sqrt{7}-10 \)

в) \( \frac{42}{2\sqrt{6}-\sqrt{3}} + \frac{24}{2\sqrt{3}+\sqrt{6}} \)

Первая дробь: \( \frac{42}{2\sqrt{6}-\sqrt{3}} = \frac{42(2\sqrt{6}+\sqrt{3})}{(2\sqrt{6}-\sqrt{3})(2\sqrt{6}+\sqrt{3})} = \frac{42(2\sqrt{6}+\sqrt{3})}{(2\sqrt{6})^2 - (\sqrt{3})^2} = \frac{42(2\sqrt{6}+\sqrt{3})}{4 \cdot 6 - 3} = \frac{42(2\sqrt{6}+\sqrt{3})}{24-3} = \frac{42(2\sqrt{6}+\sqrt{3})}{21} = 2(2\sqrt{6}+\sqrt{3}) = 4\sqrt{6}+2\sqrt{3} \)

Вторая дробь: \( \frac{24}{2\sqrt{3}+\sqrt{6}} = \frac{24(2\sqrt{3}-\sqrt{6})}{(2\sqrt{3}+\sqrt{6})(2\sqrt{3}-\sqrt{6})} = \frac{24(2\sqrt{3}-\sqrt{6})}{(2\sqrt{3})^2 - (\sqrt{6})^2} = \frac{24(2\sqrt{3}-\sqrt{6})}{4 \cdot 3 - 6} = \frac{24(2\sqrt{3}-\sqrt{6})}{12-6} = \frac{24(2\sqrt{3}-\sqrt{6})}{6} = 4(2\sqrt{3}-\sqrt{6}) = 8\sqrt{3}-4\sqrt{6} \)

Сумма: \( (4\sqrt{6}+2\sqrt{3}) + (8\sqrt{3}-4\sqrt{6}) = 4\sqrt{6}+2\sqrt{3}+8\sqrt{3}-4\sqrt{6} = 10\sqrt{3} \)

г) \( \frac{8}{\sqrt{6}+\sqrt{2}} - \frac{9}{\sqrt{6}-\sqrt{3}} \)

Первая дробь: \( \frac{8}{\sqrt{6}+\sqrt{2}} = \frac{8(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{8(\sqrt{6}-\sqrt{2})}{6-2} = \frac{8(\sqrt{6}-\sqrt{2})}{4} = 2(\sqrt{6}-\sqrt{2}) = 2\sqrt{6}-2\sqrt{2} \)

Вторая дробь: \( \frac{9}{\sqrt{6}-\sqrt{3}} = \frac{9(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})} = \frac{9(\sqrt{6}+\sqrt{3})}{6-3} = \frac{9(\sqrt{6}+\sqrt{3})}{3} = 3(\sqrt{6}+\sqrt{3}) = 3\sqrt{6}+3\sqrt{3} \)

Разность: \( (2\sqrt{6}-2\sqrt{2}) - (3\sqrt{6}+3\sqrt{3}) = 2\sqrt{6}-2\sqrt{2}-3\sqrt{6}-3\sqrt{3} = -\sqrt{6}-2\sqrt{2}-3\sqrt{3} \)

Ответ: а) 6; б) \( 4\sqrt{7}-10 \); в) \( 10\sqrt{3} \); г) \( -\sqrt{6}-2\sqrt{2}-3\sqrt{3} \).