Задание 11. Найдите значение выражения: 1 1 √5-2 √5+2'; 2 1 √10-3 - 1 10+3'; 3 1 37-6 1 37+6 1 1 5 13-3 13+3'; 6 1 √27-5 1 √27+5°

Задание 11. Найдите значение выражения:

1. $$\frac{1}{\sqrt{5}-2} - \frac{1}{\sqrt{5}+2} = \frac{\sqrt{5}+2 - (\sqrt{5}-2)}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{\sqrt{5}+2 - \sqrt{5}+2}{5-4} = \frac{4}{1} = 4$$.

2. $$\frac{1}{\sqrt{10}-3} - \frac{1}{\sqrt{10}+3} = \frac{\sqrt{10}+3 - (\sqrt{10}-3)}{(\sqrt{10}-3)(\sqrt{10}+3)} = \frac{\sqrt{10}+3 - \sqrt{10}+3}{10-9} = \frac{6}{1} = 6$$.

3. $$\frac{1}{\sqrt{37}-6} - \frac{1}{\sqrt{37}+6} = \frac{\sqrt{37}+6 - (\sqrt{37}-6)}{(\sqrt{37}-6)(\sqrt{37}+6)} = \frac{\sqrt{37}+6 - \sqrt{37}+6}{37-36} = \frac{12}{1} = 12$$.

5. $$\frac{1}{\sqrt{13}-3} - \frac{1}{\sqrt{13}+3} = \frac{\sqrt{13}+3 - (\sqrt{13}-3)}{(\sqrt{13}-3)(\sqrt{13}+3)} = \frac{\sqrt{13}+3 - \sqrt{13}+3}{13-9} = \frac{6}{4} = \frac{3}{2} = 1.5$$.

6. $$\frac{1}{\sqrt{27}-5} - \frac{1}{\sqrt{27}+5} = \frac{\sqrt{27}+5 - (\sqrt{27}-5)}{(\sqrt{27}-5)(\sqrt{27}+5)} = \frac{\sqrt{27}+5 - \sqrt{27}+5}{27-25} = \frac{10}{2} = 5$$.

Ответ: 1) 4; 2) 6; 3) 12; 5) 1.5; 6) 5