if $\tfrac{y}{2}=x$, evaluate $\left( \tfrac{{{x}^{3}}}{{{y}^{3}}}+\tfrac{1}{2} \right)\times \left( \tfrac{1}{2}-\tfrac{{{x}^{2}}}{{{y}^{2}}} \right)$
$\frac{5}{8}$
$\frac{5}{2}$
$\frac{5}{4}$
$\frac{5}{16}$
$\begin{align} & \left( \frac{{{x}^{3}}}{{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{x}^{2}}}{{{y}^{2}}} \right)=\left( \frac{{{(\tfrac{y}{2})}^{3}}}{{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{(\tfrac{y}{2})}^{2}}}{{{y}^{2}}} \right) \\ & =\left( \frac{{{y}^{3}}}{8{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{y}^{2}}}{4{{y}^{2}}} \right) \\ & =\left( \frac{1}{8}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{1}{4} \right) \\ & =\frac{5}{8}\div \frac{1}{4}=\frac{5}{8}\times \frac{4}{1}=\frac{5}{2} \end{align}$