What's the domain of $\ f(x) = \sqrt{\frac{x - 3}{x - 5}} $? What's the domain of $\ f(x) = \sqrt{\frac{x - 3}{x - 5}} $  ?
My guess is there are two possibilities depending on whether $ x - 5 $ is positive or negative, after excluding $ 5 $ of course.
 A: The term inside the square root has to be positive.Also $x\neq5$ as the denominator then becomes 0. Hence,
$$\frac{x-3}{x-5}\ge0$$
$$\Rightarrow x\in (\infty,3]\cup(5,\infty)$$
A: Assuming a function from $\mathbb{R}\to\mathbb{R}$, $\displaystyle{\frac{x-3}{x-5}}$ must be non-negative, so $x-3 \geq0$ and $x-5\geq0 \implies$ $x\geq5$. Also, $x-3\leq0$ and $x-5\leq0 \implies x\leq3$. Also, $x\neq5$ or we'll be dividing by zero.
So the domain is $\{x\in\mathbb{R}:x\leq3\;\text{or}\;x>5\}$.
If it's the complex square root function, then the domain is simply $\{x\in\mathbb{R}:x\neq5\}$.
A: HINT:
The thing under the root must be positive. That happens when both numerator and denominator are positive, or when both are negative.
A: You seem to know that the inside of the square root must be non-negative. In your case,
(1) $\dfrac{x-3}{x-5} \geq 0$.
Moreover, as you've found, the denominator can not be zero. Namely 
(2) $x-5 \ne 0$.
Under the condition, let's consider the condition (1).
$$\frac{x-3}{x-5} \geq 0 \iff \frac{x-3}{x-5}(x-5)^2 \geq 0.$$
For the equivalence, I used the fact that the square of a real number is always non-negative. I think that you can simplify the LHS of the last inequality and reach the answer.
