Answer
$\dfrac{1}{x(x+3)}$
Work Step by Step
The given expression, $
\dfrac{2x^2+7}{2x^4-18x^2}-\dfrac{6x+7}{2x^4-18x^2}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{2x^2+7-(6x+7)}{2x^4-18x^2}
\\\\=
\dfrac{2x^2+7-6x-7}{2x^4-18x^2}
\\\\=
\dfrac{2x^2-6x}{2x^4-18x^2}
\\\\=
\dfrac{2x(x-3)}{2x^2(x^2-9)}
\\\\=
\dfrac{2x(x-3)}{2x^2(x+3)(x-3)}
\\\\=
\dfrac{\cancel{2x}(\cancel{x-3})}{\cancel{2x}\cdot x(x+3)(\cancel{x-3})}
\\\\=
\dfrac{1}{x(x+3)}
.\end{array}
