Answer
$\{x|x\geq 3\}$
Work Step by Step
We know that $\displaystyle \frac{\sqrt{x-3}}{x^{2}-4x+4}$ is undefined whenever the denominator is 0, so:
$x^2-4x+4=(x-2)^2=0$
$x=2$
We also know that the equation is undefined for negative roots:
$x-3<0$
$x<3$
So $x$ must be greater than or equal to 3 (which also ensures that it's not 2). Thus the domain is:
$\{x|x\geq 3\}$