Задание 10. Найдите значение выражения: 1 1 2+√3 + 1 2-√3'; 2 1 5+√23 + 1 5-√23 3 1 6+√35 + 1 6-35' 1 + 1 4 4+15+4-15' 1 + 1 57+√47 7-√47' 6 1 1 3+√7 3-√7

Задание 10. Найдите значение выражения:

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

2. $$\frac{1}{5+\sqrt{23}} + \frac{1}{5-\sqrt{23}} = \frac{5-\sqrt{23} + 5 + \sqrt{23}}{(5+\sqrt{23})(5-\sqrt{23})} = \frac{10}{25-23} = \frac{10}{2} = 5$$.

3. $$\frac{1}{6+\sqrt{35}} + \frac{1}{6-\sqrt{35}} = \frac{6-\sqrt{35} + 6 + \sqrt{35}}{(6+\sqrt{35})(6-\sqrt{35})} = \frac{12}{36-35} = \frac{12}{1} = 12$$.

4. $$\frac{1}{4+\sqrt{15}} + \frac{1}{4-\sqrt{15}} = \frac{4-\sqrt{15} + 4 + \sqrt{15}}{(4+\sqrt{15})(4-\sqrt{15})} = \frac{8}{16-15} = \frac{8}{1} = 8$$.

5. $$\frac{1}{7+\sqrt{47}} + \frac{1}{7-\sqrt{47}} = \frac{7-\sqrt{47} + 7 + \sqrt{47}}{(7+\sqrt{47})(7-\sqrt{47})} = \frac{14}{49-47} = \frac{14}{2} = 7$$.

6. $$\frac{1}{3+\sqrt{7}} + \frac{1}{3-\sqrt{7}} = \frac{3-\sqrt{7} + 3 + \sqrt{7}}{(3+\sqrt{7})(3-\sqrt{7})} = \frac{6}{9-7} = \frac{6}{2} = 3$$.

Ответ: 1) 4; 2) 5; 3) 12; 4) 8; 5) 7; 6) 3