#### Answer

Choice H

#### Work Step by Step

Using $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x,$ then
\begin{array}{l}\require{cancel}
x^{-2/5}
\\\\=
\dfrac{1}{x^{2/5}}
.\end{array}
Using $x^{m/n}=\sqrt[n]{x^m}=\left(\sqrt[n]{x} \right)^m,$ then
\begin{array}{l}\require{cancel}
\dfrac{1}{x^{2/5}}
\\\\=
\dfrac{1}{(\sqrt[5]{x})^2}
.\end{array}
Hence, $\text{
Choice H
.}$