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The task is to find the domain of the given functions:

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

Let's determine the domain for each function:

### 1. Function: \( f(x) = \frac{x + 3}{x - 4} \)

The domain of a rational function excludes values that make the denominator zero. For this function, we need to find when the denominator is zero:

\[
x - 4 = 0 \implies x = 4
\]

Thus, the domain is all real numbers except \( x = 4 \). 

**Domain:** \( x \in \mathbb{R}, x \neq 4 \).

### 2. Function: \( f(x) = \sqrt{x - 1} + \sqrt{x - 3} \)

The domain of this function depends on the values inside the square roots being non-negative. We set up the conditions:

1. For \( \sqrt{x - 1} \): 
   \[
   x - 1 \geq 0 \implies x \geq 1
   \]

2. For \(

Знайдіть область визначення функції:
1) f(x) = \frac{x + 3}{x - 4};
4) f(x) = \sqrt{x - 1} + \sqrt{x - 3}

Learn how to find the domain of rational and square root functions. Solve f(x) = (x + 3)/(x - 4) and f(x) = √(x - 1) + √(x - 3) with step-by-step solutions. This problem covers key concepts like rational and square root functions, suitable for Grades 9-11.

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