$$\frac{(\frac{1}{2}\rfloor^{-8}\cdot(\frac{2}{3})^{-2}\cdot(\frac{9}{2})^{3}}{2^{3}\cdot(\frac{4}{9})^{-1}\cdot(\frac{1}{3})^{-3}\cdot(\frac{4}{3})^{2}}=$$
