The question asks to find the domain of the following functions:

(a) $f(x) = \frac{2x + 1}{x^2 + x - 2}$

We need to find the values of $x$ for which the denominator is not zero. The domain will be all real values except those that make the denominator zero.

The denominator is: $x^2 + x - 2 = 0$ Factoring: $(x - 1)(x + 2) = 0$ So, $x = 1$ and $x = -2$ are the values where the denominator is zero. Thus, the domain of $f(x)$ is: $\boxed{(-\infty, -2) \cup (-2, 1) \cup (1, \infty)}$

(b) $g(x) = \frac{\sqrt{x}}{x + 1}$

For the square root function $\sqrt{x}$ to be defined, $x$ must be non-negative, meaning $x \geq 0$. Also, the denominator cannot be zero, so $x + 1 \neq 0$, which gives $x \neq -1$.

Thus, the domain of $g(x)$ is: $\boxed{[0, \infty) \setminus \{-1\}}$ This means all non-negative real numbers excluding $-1$.

(c) $h(x) = \sqrt{x - 1} + \sqrt{x - 1}$

This function contains square roots, so the expressions inside the square roots must be non-negative. Both square root terms involve $x - 1$, so we must have: $x - 1 \geq 0 \quad \Rightarrow \quad x \geq 1$ Thus, the domain of $h(x)$ is: $\boxed{[1, \infty)}$

Would you like any further details or clarification on these calculations?

Here are 5 related questions to consider:

1. How do you determine the domain of a rational function in general?
2. What happens when there are square roots in the denominator?
3. What is the domain of $f(x) = \frac{1}{x^2 + 4}$?
4. How do restrictions in square roots affect the domain?
5. How would the domain change if the function were $f(x) = \sqrt{1 - x}$?

Tip: Always check for restrictions such as division by zero or square roots of negative numbers when determining a function's domain.