Which of the following properly factors the expression to prepare for reducing? (A) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-3)(x-1)}{(x-2)(x-2)}$ (B) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-3)(x-1)}{(x-2)(x+2)}$ (C) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-1)(x-2)}{(x-2)(x-2)}$ (D) $\frac{x^{2}-3 x+2}{(x-2)^{2}}=\frac{(x-1)(x-2)}{(x-2)(x+2)}$

Step 1 ：The problem is asking to factorize the expression \( \frac{x^{2}-3 x+2}{(x-2)^{2}} \).

Step 2 ：The numerator \( x^{2}-3 x+2 \) can be factorized into \( (x-1)(x-2) \).

Step 3 ：The denominator is already in the factorized form \( (x-2)^{2} \).

Step 4 ：So, the correct factorization of the given expression is \( \frac{(x-1)(x-2)}{(x-2)(x-2)} \).

Step 5 ：Checking the options, we find that this matches with option (C).

Step 6 ：Final Answer: \(\boxed{(C)}\)