Answer
$\dfrac{5\sqrt{7}}{2x}$
Work Step by Step
Using the properties of radicals, then,
\begin{array}{l}
\sqrt{\dfrac{28}{x^2}}+\sqrt{\dfrac{7}{4x^2}}
\\\\=
\sqrt{\dfrac{4}{x^2}\cdot7}+\sqrt{\dfrac{1}{4x^2}\cdot7}
\\\\=
\dfrac{2\sqrt{7}}{x}+\dfrac{\sqrt{7}}{2x}
\\\\=
\dfrac{2(2\sqrt{7})+\sqrt{7}}{2x}
\\\\=
\dfrac{4\sqrt{7}+\sqrt{7}}{2x}
\\\\=
\dfrac{5\sqrt{7}}{2x}
.\end{array}