#### Answer

$\dfrac{3}{2x}$

#### Work Step by Step

RECALL:
$x^{-m} = \dfrac{1}{x^{m}}, m \gt 0$
Use the rule above to have:
$\\=\dfrac{1}{x^1} + \frac{1}{(2x)^1}
\\=\dfrac{1}{x}+\dfrac{1}{2x}$
Make the expressions similar using the LCD which is $2x$ to have:
$\\=\dfrac{1(2)}{2(x)} + \dfrac{1}{2x}
\\=\dfrac{2}{2x} + \dfrac{1}{2x}
\\=\dfrac{3}{2x}$