Answer
$\dfrac{7\sqrt{11}+7\sqrt{7}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the given expression, $
\dfrac{14}{\sqrt{11}-\sqrt{7}}
,$ multiply the numerator and the denominator by the conjugate of the denominator.
$\bf{\text{Solution Details:}}$
Reversing the operator between the expression, $
\sqrt{11}-\sqrt{7}
,$ then the conjugate is $
\sqrt{11}+\sqrt{7}
.$ Multiplying both the numerator and the denominator by the conjugate results to
\begin{array}{l}\require{cancel}
\dfrac{14}{\sqrt{11}-\sqrt{7}}\cdot\dfrac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}
\\\\=
\dfrac{14(\sqrt{11}+\sqrt{7})}{(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})}
.\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{14(\sqrt{11}+\sqrt{7})}{(\sqrt{11})^2-(\sqrt{7})^2}
\\\\=
\dfrac{14(\sqrt{11}+\sqrt{7})}{11-7}
\\\\=
\dfrac{14(\sqrt{11}+\sqrt{7})}{4}
\\\\=
\dfrac{\cancel2(7)(\sqrt{11}+\sqrt{7})}{\cancel2(2)}
\\\\=
\dfrac{7(\sqrt{11}+\sqrt{7})}{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{7\sqrt{11}+7\sqrt{7}}{2}
.\end{array}
