206.1) f(x) = 6√x² + 1 / x, x = 2√2; 2) f(x) = √x² - 1 / x, x = √5; 3) f(x) = 2x + 1 / √x² + 1, x = √3.

Решение:

1. Для \( f(x) = \frac{6\sqrt{x^2 + 1}}{x} \) при \( x = 2\sqrt{2} \): \( f(2\sqrt{2}) = \frac{6\sqrt{(2\sqrt{2})^2 + 1}}{2\sqrt{2}} = \frac{6\sqrt{8 + 1}}{2\sqrt{2}} = \frac{6\sqrt{9}}{2\sqrt{2}} = \frac{6 \cdot 3}{2\sqrt{2}} = \frac{18}{2\sqrt{2}} = \frac{9}{\sqrt{2}} = \frac{9\sqrt{2}}{2} \).
2. Для \( f(x) = \frac{\sqrt{x^2 - 1}}{x} \) при \( x = \sqrt{5} \): \( f(\sqrt{5}) = \frac{\sqrt{(\sqrt{5})^2 - 1}}{\sqrt{5}} = \frac{\sqrt{5 - 1}}{\sqrt{5}} = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \).
3. Для \( f(x) = \frac{2x + 1}{\sqrt{x^2 + 1}} \) при \( x = \sqrt{3} \): \( f(\sqrt{3}) = \frac{2\sqrt{3} + 1}{\sqrt{(\sqrt{3})^2 + 1}} = \frac{2\sqrt{3} + 1}{\sqrt{3 + 1}} = \frac{2\sqrt{3} + 1}{\sqrt{4}} = \frac{2\sqrt{3} + 1}{2} = \sqrt{3} + \frac{1}{2} \).

Ответ: Для 1) \(f(2\sqrt{2}) = \frac{9\sqrt{2}}{2}\); для 2) \(f(\sqrt{5}) = \frac{2\sqrt{5}}{5}\); для 3) \(f(\sqrt{3}) = \sqrt{3} + \frac{1}{2}\).