Answer
$3+2\sqrt{2}$
Work Step by Step
Multiplying the given expression $
\dfrac{\sqrt{2}+1}{\sqrt{2}-1}
$ by the conjugate of the denominator, then the rationalized denominator form is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\cdot\dfrac{\sqrt{2}+1}{\sqrt{2}+1}
\\\\=
\dfrac{(\sqrt{2}+1)^2}{(\sqrt{2})^2-(1)^2}
\\\\=
\dfrac{(\sqrt{2})^2+2(\sqrt{2})(1)+1^2}{2-1}
\\\\=
\dfrac{2+2\sqrt{2}+1}{1}
\\\\=
3+2\sqrt{2}
.\end{array}
