Answer
$\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}$
Work Step by Step
Multiplying by the conjugate of the numerator, the rationalized (numerator) form of the given expression, $
\dfrac{5+\sqrt{2}}{\sqrt{2x}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{5+\sqrt{2}}{\sqrt{2x}}\cdot\dfrac{5-\sqrt{2}}{5-\sqrt{2}}
\\=
\dfrac{(5+\sqrt{2})(5-\sqrt{2})}{\sqrt{2x}(5-\sqrt{2})}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{(5)^2-(\sqrt{2})^2}{\sqrt{2x}(5-\sqrt{2})}
\\=
\dfrac{25-2}{\sqrt{2x}(5-\sqrt{2})}
\\=
\dfrac{23}{\sqrt{2x}(5-\sqrt{2})}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{23}{\sqrt{2x}(5)+\sqrt{2x}(-\sqrt{2})}
\\=
\dfrac{23}{5\sqrt{2x}-\sqrt{2x(2)}}
\\=
\dfrac{23}{5\sqrt{2x}-\sqrt{4x}}
\\=
\dfrac{23}{5\sqrt{2x}-2\sqrt{x}}
.\end{array}
