Answer
$1$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, then the given expression, $
\dfrac{x^2-3x}{x^3-27}\div\dfrac{2x}{2x^2+6x+18}
$, simplifies to
\begin{array}{l}\require{cancel}
\dfrac{x^2-3x}{x^3-27}\cdot\dfrac{2x^2+6x+18}{2x}
\\\\=
\dfrac{x(x-3)}{(x-3)(x^2+3x+9)}\cdot\dfrac{2(x^2+3x+9)}{2x}
\\\\=
\dfrac{\cancel{x}(\cancel{x-3})}{(\cancel{x-3})(\cancel{x^2+3x+9})}\cdot\dfrac{\cancel{2}(\cancel{x^2+3x+9})}{\cancel{2}\cancel{x}}
\\\\=
1
.\end{array}
