Notice that the function does not hold for values of x that cancel hence, you need to eliminate the zeroes of such that:

Notice that the domain does not contain the value .

There exists a second condition to be considered for the square root to exist:

**Hence, evaluating the domain of the function yields **

The denominator of the function impose the domain of definition.

We'll impose the constraint of existence of the function:

5x-125 > 0

We've imposed that the expression to be strictly positive, since the sqrt 5x-125 represents the denominator. According to the rule, the values of denominator have to be other than zero.

We'll divide by 5:

x - 25>0

We'll add 25 both sides:

x > 25

The domain of definition is the interval of x values, that are strictly greater than 25!

x belongs to (25, +infinite)