Answer
$\dfrac{2}{5}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $
\dfrac{x^2-6x+9}{2x^2-18}\cdot \dfrac{4x+12}{5x-15}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{(x-3)(x-3)}{2(x^2-9)}\cdot \dfrac{4(x+3)}{5(x-3)}
\\\\=
\dfrac{(x-3)(x-3)}{2(x+3)(x-3)}\cdot \dfrac{4(x+3)}{5(x-3)}
\\\\=
\dfrac{(\cancel{x-3})(\cancel{x-3})}{\cancel{2}(\cancel{x+3})(\cancel{x-3})}\cdot \dfrac{\cancel{2}\cdot2(\cancel{x+3})}{5(\cancel{x-3})}
\\\\=
\dfrac{2}{5}
.\end{array}
