Answer:
The expression is unclear so:
If the denominator is [tex]\sqrt{x}-1[/tex] then your set of existence is given by
[tex]\left \{ {{x\ge0} \atop {x\ne1}} \right.[/tex]
since you want the quantity inside both square roots to be positive (and it happens for both to be simply [tex]x\ge0[/tex] and you don't want a zero at the denominator so you have to rule out 1.
If the square root includes the whole denominator [tex]\sqrt{x-1}[/tex] the condition becomes [tex]\left \{ {{x\ge0} \atop {x>1}} \right.[/tex]
where you don't include the extreme in the second condition since, again, you don't want to divide by 0. The answer in this case is simply [tex]x>1[/tex] since values between 0 and 1 will give a negative root which is not a real number.
