Answer
$\dfrac{(x-3)^2}{x-2}$
Work Step by Step
The given expression, $ \dfrac{\dfrac{x^2-5x+6}{x+3}}{\dfrac{x^2-4x+4}{x^2-9}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{x^2-5x+6}{x+3}\div\dfrac{x^2-4x+4}{x^2-9} \\\\= \dfrac{x^2-5x+6}{x+3}\cdot\dfrac{x^2-9}{x^2-4x+4} \\\\=
\dfrac{(x-2)(x-3)}{x+3}\cdot\dfrac{(x+3)(x-3)}{(x-2)(x-2)}
\\\\=
\dfrac{(\cancel{x-2})(x-3)}{\cancel{x+3}}\cdot\dfrac{(\cancel{x+3})(x-3)}{(\cancel{x-2})(x-2)}
\\\\=
\dfrac{(x-3)^2}{x-2}
.\end{array}
