$?解：原式 ={\frac {(a-2)(a+2)-a(a-1)} {a{(a+2)}^{2}}}·{\frac {a+2} {a-4}}$
$ ={\frac {({a}^{2}-4)-({a}^{2}-a)} {a{(a+2)}^{2}}}·{\frac {a+2} {a-4}}$
$ ={\frac {a-4} {a{(a+2)}^{2}}}·{\frac {a+2} {a-4}}$
$ ={\frac {1} {{a}^{2}+2a}}$
$∵ {a}^{2}+2a-3=0$
$∴ {a}^{2}+2a=3$
$將{a}^{2}+2a=3代入{\frac {1} {{a}^{2}+2a}}，得$
$原式 ={\frac {1} {3}}$
