The problem asks us to find the domain of the given functions. We need to consider cases where we have square roots (the expression inside must be non-negative) and fractions (the denominator cannot be zero).

1. Problem Description

The problem asks us to find the domain of the given functions. We need to consider cases where we have square roots (the expression inside must be non-negative) and fractions (the denominator cannot be zero).

2. Solution Steps

1. $f(x) = \frac{\sqrt{2x-10}}{x-7}$

We need $2x-10 \ge 0$ and $x-7 \ne 0$.

$2x \ge 10 \implies x \ge 5$.

$x \ne 7$.

So, the domain is $[5, 7) \cup (7, \infty)$.

2. $f(x) = \frac{2}{x^2 + 3x}$

We need $x^2 + 3x \ne 0$.

$x(x+3) \ne 0 \implies x \ne 0$ and $x \ne -3$.

So, the domain is $(-\infty, -3) \cup (-3, 0) \cup (0, \infty)$.

3. $f(x) = \frac{4x-1}{\sqrt{5-2x}}$

We need $5-2x > 0$.

$5 > 2x \implies x < \frac{5}{2}$.

So, the domain is $(-\infty, \frac{5}{2})$.

4. $f(x) = \sqrt{\frac{3x-1}{x+4}}$

We need $\frac{3x-1}{x+4} \ge 0$.

Case 1: $3x-1 \ge 0$ and $x+4 > 0$.

$x \ge \frac{1}{3}$ and $x > -4$. Thus $x \ge \frac{1}{3}$.

Case 2: $3x-1 \le 0$ and $x+4 < 0$.

$x \le \frac{1}{3}$ and $x < -4$. Thus $x < -4$.

So, the domain is $(-\infty, -4) \cup [\frac{1}{3}, \infty)$.

5. $f(x) = \frac{\sqrt{3x-1}}{\sqrt{x+4}}$

We need $3x-1 \ge 0$ and $x+4 > 0$.

$3x \ge 1 \implies x \ge \frac{1}{3}$.

$x > -4$.

So, the domain is $[\frac{1}{3}, \infty)$.

6. $f(x) = \sqrt{x^2 - 11x + 18}$

We need $x^2 - 11x + 18 \ge 0$.

$x^2 - 11x + 18 = (x-2)(x-9)$.

So we need $(x-2)(x-9) \ge 0$.

Case 1: $x-2 \ge 0$ and $x-9 \ge 0$.

$x \ge 2$ and $x \ge 9$. Thus $x \ge 9$.

Case 2: $x-2 \le 0$ and $x-9 \le 0$.

$x \le 2$ and $x \le 9$. Thus $x \le 2$.

So, the domain is $(-\infty, 2] \cup [9, \infty)$.

7. $f(x) = \frac{2+x}{\sqrt{4x-1}}$

We need $4x-1 > 0$.

$4x > 1 \implies x > \frac{1}{4}$.

So, the domain is $(\frac{1}{4}, \infty)$.

8. $f(x) = \sqrt{\frac{x^2 - 25}{8-x}}$

We need $\frac{x^2 - 25}{8-x} \ge 0$.

$\frac{(x-5)(x+5)}{8-x} \ge 0$.

Case 1: $(x-5)(x+5) \ge 0$ and $8-x > 0$.

$x \le -5$ or $x \ge 5$, and $x < 8$. Thus $x \in (-\infty, -5] \cup [5, 8)$.

Case 2: $(x-5)(x+5) \le 0$ and $8-x < 0$.

$-5 \le x \le 5$ and $x > 8$. No solution.

So, the domain is $(-\infty, -5] \cup [5, 8)$.

9. $f(x) = \frac{1}{x^2 + 2x + 5}$

We need $x^2 + 2x + 5 \ne 0$.

$x^2 + 2x + 5 = (x+1)^2 + 4$. Since $(x+1)^2 \ge 0$, $(x+1)^2 + 4 \ge 4 > 0$.

So the denominator is always non-zero.

So, the domain is $(-\infty, \infty)$.

1

0. $f(x) = \frac{\sqrt{x+3}}{\sqrt{x^2-4}}$

We need $x+3 \ge 0$ and $x^2-4 > 0$.

$x \ge -3$ and $(x-2)(x+2) > 0$.

$x \ge -3$ and ($x > 2$ or $x < -2$).

Combining these, we have $[-3, -2) \cup (2, \infty)$.

3. Final Answer

1. $[5, 7) \cup (7, \infty)$

2. $(-\infty, -3) \cup (-3, 0) \cup (0, \infty)$

3. $(-\infty, \frac{5}{2})$

4. $(-\infty, -4) \cup [\frac{1}{3}, \infty)$

5. $[\frac{1}{3}, \infty)$

6. $(-\infty, 2] \cup [9, \infty)$

7. $(\frac{1}{4}, \infty)$

8. $(-\infty, -5] \cup [5, 8)$

9. $(-\infty, \infty)$

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0. $[-3, -2) \cup (2, \infty)$