Answer
$\dfrac{-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the denominator of the given expression, $
\dfrac{\sqrt{7}-1}{2\sqrt{7}+4\sqrt{2}}
,$ multiply the numerator and the denominator by the conjugate of the denominator.
$\bf{\text{Solution Details:}}$
Reversing the operator between the terms of the denominator in the expression above, the conjugate of the denominator is $
2\sqrt{7}-4\sqrt{2}
.$ Multiplying the conjugate to both the numerator and the denominator of the expression above results to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{7}-1}{2\sqrt{7}+4\sqrt{2}}\cdot\dfrac{2\sqrt{7}-4\sqrt{2}}{2\sqrt{7}-4\sqrt{2}}
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\dfrac{\sqrt{7}(2\sqrt{7})+\sqrt{7}(-4\sqrt{2})-1(2\sqrt{7})-1(-4\sqrt{2})}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})}
.\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{2\sqrt{7(7)}-4\sqrt{7(2)}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})}
\\\\=
\dfrac{2\sqrt{7^2}-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})}
\\\\=
\dfrac{2(7)-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\sqrt{2})}
\\\\=
\dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7}+4\sqrt{2})(2\sqrt{7}-4\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{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{(2\sqrt{7})^2-(4\sqrt{2})^2}
\\\\=
\dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{4(7)-16(2)}
\\\\=
\dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{28-32}
\\\\=
\dfrac{14-4\sqrt{14}-2\sqrt{7}+4\sqrt{2}}{-4}
\\\\=
\dfrac{-2(-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2})}{-4}
\\\\=
\dfrac{\cancel{-2}(-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2})}{\cancel{-4}^2}
\\\\=
\dfrac{-7+2\sqrt{14}+\sqrt{7}-2\sqrt{2}}{2}
.\end{array}
