((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

How would I solve this?

((x-2)/(2x)+(1)/(x+2))/((3)/(2)-(6)/(x^(2)+2x))

$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x^{2}+2x} } } \quad & | \quad x^{2}+2x = x(x+2) \\\\ &=& \dfrac{ \dfrac{x-2}{2x} + \dfrac{1}{x+2} } { \dfrac{3}{2}- \dfrac{6} {x(x+2)} } \\\\ &=& \dfrac{ \dfrac{(x-2)(x+2) + 1\cdot 2x}{2x(x+2)} } { \dfrac{3x(x+2)-2\cdot 6}{2x(x+2)} } \\\\ &=& \dfrac{\left[~(x-2)(x+2) + 1\cdot 2x ~\right]}{2x(x+2)} \cdot \dfrac{2x(x+2)} {\left[~ 3x(x+2)-2\cdot 6 ~\right] } \\\\ &=& \dfrac{(x-2)(x+2) + 1\cdot 2x} {3x(x+2)-2\cdot 6} \\\\ &=& \dfrac{(x-2)(x+2) + 2x} {3x(x+2)-12} \\\\ &=& \dfrac{x^2-4 + 2x } {3x^2+6x-12} \\\\ &=& \dfrac{1}{3} \cdot \dfrac{\left(x^2+2x-4\right)} {\left(x^2+2x-4\right)} \\\\ &\mathbf{=}& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array}$