#### Answer

$-3+2\sqrt{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{1-\sqrt{2}}{1+\sqrt{2}}
,$ 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 by the conjugate of the denominator results to
\begin{array}{l}\require{cancel}
\dfrac{1-\sqrt{2}}{1+\sqrt{2}}\cdot\dfrac{1-\sqrt{2}}{1-\sqrt{2}}
\\\\=
\dfrac{(1-\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}
\\\\=
\dfrac{(1-\sqrt{2})^2}{(1+\sqrt{2})(1-\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{(1-\sqrt{2})^2}{(1)^2-(\sqrt{2})^2}
\\\\=
\dfrac{(1-\sqrt{2})^2}{1-2}
\\\\=
\dfrac{(1-\sqrt{2})^2}{-1}
\\\\=
-(1-\sqrt{2})^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-[(1)^2-2(1)(\sqrt{2})+(\sqrt{2})^2]
\\\\=
-[1-2\sqrt{2}+2]
\\\\=
-[3-2\sqrt{2}]
\\\\=
-3+2\sqrt{2}
.\end{array}