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

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**For the first function:**  
$$
f(x)=\frac{x^{2}-4x}{x^{2}-3x-4}
$$

1. The only possible issues occur when the denominator is zero. Set the denominator equal to zero:
   $$
   x^{2}-3x-4=0
   $$
2. Factor the quadratic:
   $$
   (x-4)(x+1)=0
   $$
3. Solve for $$x$$:
   $$
   x-4=0 \quad \Rightarrow \quad x=4
   $$
   $$
   x+1=0 \quad \Rightarrow \quad x=-1
   $$
4. The values $$x=4$$ and $$x=-1$$ must be excluded from the domain.

Thus, the natural domain is:
$$
\{x\in\mathbb{R} : x \neq 4 \text{ and } x \neq -1\}
$$

---

**For the second function:**  
$$
f(x)=\frac{1}{x^{2}-25}+\sqrt{x^{2}-4x+3}
$$

1. **For the Fraction Part:**

The denominator cannot be zero. Set:
   $$
   x^{2}-25=0
   $$
   Solve for $$x$$:
   $$
   x^{2}=25 \quad \Rightarrow \quad x=5 \quad \text{or} \quad x=-5
   $$
   So, $$x$$ cannot be $$5$$ or $$-5$$.

2. **For the Square Root Part:**

The expression under the square root must be non-negative:
   $$
   x^{2}-4x+3\ge0
   $$
3. Factor the quadratic:
   $$
   (x-1)(x-3)\ge0
   $$
4. Determine the intervals where the inequality holds. The zeros are $$x=1$$ and $$x=3$$. Analyzing the sign:
   - For $$x<1$$: both factors are negative, so their product is positive.
   - For $$13$$: both factors are positive, so the product is positive.

Hence, the inequality holds when:
   $$
   x\le1 \quad \text{or} \quad x\ge3
   $$

5. Combine the two pieces of information, remembering to exclude $$x=-5$$ from the first interval and $$x=5$$ from the second interval.

Thus, the natural domain is:
$$
\left((-\infty,1]\setminus\{-5\}\right)\cup\left([3,\infty)\setminus\{5\}\right)
$$
**Natural Domain:** 1. **First Function:** - Exclude $$x = 4$$ and $$x = -1$$. - Domain: All real numbers except $$x = 4$$ and $$x = -1$$. 2. **Second Function:** - Exclude $$x = 5$$ and $$x = -5$$. - Also, include all real numbers where $$x \leq 1$$ or $$x \geq 3$$. - Domain: $$x \leq 1$$ (excluding $$x = -5$$) and $$x \geq 3$$ (excluding $$x = 5$$).

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

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