d the natural domain of \[ \begin{array}{l}f(x)=\frac{x^{2}-4 x}{x^{2}-3 x-4} \\ f(x)=\frac{1}{x^{2}-25}+\sqrt{x^{2}-4 x+3}\end{array} \]

To determine the natural domain of the functions provided, we need to identify any values for \( x \) that would cause the expressions to be undefined. For the first function \( f(x) = \frac{x^{2} - 4x}{x^{2} - 3x - 4} \), we find the values that make the denominator zero. Setting \( x^{2} - 3x - 4 = 0 \) and factoring gives us \( (x - 4)(x + 1) = 0 \). Thus, \( x = 4 \) and \( x = -1 \) are excluded from the domain. Therefore, the natural domain is all real numbers except for \( x = 4 \) and \( x = -1 \). For the second function \( f(x) = \frac{1}{x^{2} - 25} + \sqrt{x^{2} - 4x + 3} \), we first address the denominator. The expression \( x^{2} - 25 = 0 \) factors to \( (x - 5)(x + 5) = 0 \), which means \( x = 5 \) and \( x = -5 \) are problematic. Next, we ensure the argument of the square root \( x^{2} - 4x + 3 \geq 0 \). Factoring yields \( (x - 1)(x - 3) \geq 0 \), which holds for \( x \leq 1 \) or \( x \geq 3 \). Combining these conditions, we conclude that the natural domain excludes \( x = 5 \) and \( x = -5 \), while also being restricted to \( x \leq 1 \) or \( x \geq 3 \). In summary: - For the first function, the domain is \( \mathbb{R} \setminus \{-1, 4\} \). - For the second function, the domain is \( (-\infty, -5) \cup (-5, 1] \cup [3, 5) \cup (5, \infty) \).