1) $\frac{3^{5 / 2}}{3^{2 / 3}}=\ldots$
(A) $\frac{1}{3}$
$\begin{array}{lll}\text { (B) } 3 & \text { (C) } 3^{15 / 4}\end{array}$
(D) $3^{11 / 6}$
(E) $3^{19 / 6}$.
1) The answer is $(\mathbf{D})$. Indeed $\frac{3^{5 / 2}}{3^{2 / 3}}=3^{5 / 2} 3^{-2 / 3}=3^{5 / 2-2 / 3}=3^{11 / 6}$.