205.1) f(x) = x√x² + 1, x = √3; 2) f(x) = (x² - 1) √x² - 1, x = √2; 3) f(x) = (x² + 1) √x² + 1, x = √3.

Решение:

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

Ответ: Для 1) \(f(\sqrt{3}) = 2\sqrt{3}\); для 2) \(f(\sqrt{2}) = 1\); для 3) \(f(\sqrt{3}) = 8\).