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)