第93問 答え

$$\begin{align}& \frac{2}{3} x^2y^3 \div \left(-\frac{1}{8} xy\right) \div \frac{4}{9}y\\&=\frac{2}{3}x^2y^3 \times \left(-\frac{8}{xy}\right) \times \frac{9}{4y}\\&=-\frac{2 x^2y^3 \times 8 \times 9}{3 \times xy \times 4y}\\&=-12xy\end{align}$$