207.1) f(x) = 9x / √x² + 1, x = 2√2; 2) f(x) = 4x / √x² - 1, x = √5; 3) f(x) = x² / √x² + 1, x = √3.

Решение:

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

Ответ: Для 1) \(f(2\sqrt{2}) = 6\sqrt{2}\); для 2) \(f(\sqrt{5}) = 2\sqrt{5}\); для 3) \(f(\sqrt{3}) = \frac{3}{2}\).