Answer
(a) $x^{5/4}$
(b) $\frac{4}{\sqrt[3]{y}}$
Work Step by Step
(a) $\displaystyle \frac{x^{3/4_{X}7/4}}{x^{5/4}}=x^{3/4+7/4-5/4}=x^{5/4}$
(b)$\displaystyle \frac{(2y^{4/3})^{2}y^{-2/3}}{y^{7/3}}=4y^{8/3-2/3-7/3}=4y^{-1/3}=\frac{4}{y^{1/3}}=\frac{4}{\sqrt[3]{y}}$
