Another Radical Expression Question

This is my solution

If $2<x<3$ , then $\sqrt(x-2)^2=x-2$ while $|x-3|=-(x-3)=3-x$ such that we have $x-2+3-x=\boxed{1}$ where the x's actually cancel out each other.

since,2<x<3,i.e,x=[2.1,2.9]. so, sqrt( (x-2)^2 ) = |x-2| = [0.1 , 0.9]

&&

|x-3| = [0.1 , 0.9] or, simply [0.9 , 0.1]

Hence,sum i.e., ? = 1.

Note: sqrt( X^2 ) = |X|.