$解：原式 ={\frac {(x+y)-(x-y)} {(x+y)(x-y)}}÷{\frac {2y} {{(x+y)}^{2}}}$
$\ \ \ \ \ \ \ \ ={\frac {2y} {(x+y)(x-y)}}·{\frac {{(x+y)}^{2}} {2y}}$
$\ \ \ \ \ \ \ \ ={\frac {x+y} {x-y}}$
$將x=\sqrt {3}+\sqrt {2}，y=\sqrt {3}-\sqrt {2}代入，得$
$原式 ={\frac {({\sqrt {3}+{\sqrt {2}}})+({\sqrt {3}-{\sqrt {2}}})} {({\sqrt {3}+{\sqrt {2}}})-({\sqrt {3}-{\sqrt {2}}})}}$
$\ \ \ \ ={\frac {2\sqrt {3}} {2\sqrt {2}}}$
$\ \ \ \ ={\frac {\sqrt {6}} {2}}$
