The expression$\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified is:
$\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$
Factoring each of the binomials in the expression $\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ will yield the result of \[\frac{(x-2)(x-1)}{(x-3)(x-2)}\div \frac{(x-4)(x-1)}{(x-3)(x-4)},\]
We can eliminate like terms to get $\frac {x-1}{x-3}\div \frac{x-1}{x-3}$, which, according to identity property, is equivalent to the answer (D) 1.