#### Answer

$\dfrac{10k-5\sqrt{kq}}{4k-q}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{5\sqrt{k}}{2\sqrt{k}+\sqrt{q}}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5\sqrt{k}}{2\sqrt{k}+\sqrt{q}}\cdot\dfrac{2\sqrt{k}-\sqrt{q}}{2\sqrt{k}-\sqrt{q}}
\\\\=
\dfrac{5\sqrt{k}(2\sqrt{k}-\sqrt{q})}{(2\sqrt{k}+\sqrt{q})(2\sqrt{k}-\sqrt{q})}
.\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\sqrt{k}(2\sqrt{k}-\sqrt{q})}{(2\sqrt{k})^2-(\sqrt{q})^2}
\\\\=
\dfrac{5\sqrt{k}(2\sqrt{k}-\sqrt{q})}{4k-q}
.\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{5(2)(\sqrt{k})^2-5(\sqrt{k})(\sqrt{q})}{4k-q}
\\\\=
\dfrac{10k-5(\sqrt{k})(\sqrt{q})}{4k-q}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{10k-5\sqrt{k(q)}}{4k-q}
\\\\=
\dfrac{10k-5\sqrt{kq}}{4k-q}
.\end{array}