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