2≤x≤3
x-2≥0 x-3≤0
|x-2|=x-2 |x-3|=3-x
√(x²-4x+4)+|x-3|
=√(x-2)²+|x-3|
=|x-2|+|x-3|
=x-2+(3-x)
=1
选D.