解:原式$=\frac {a²}{a(a-3)}-\frac {a+6}{a²-3a}+\frac {3(a-3)}{a(a-3)}$
$=\frac {a²-a-6+3a-9}{a(a-3)}$
$=\frac {a²+2a-15}{a(a-3)}$
$=\frac {(a-3)(a+5)}{a(a-3)}$
$=\frac {a+5}{a}$
把$a=\frac {3}{2}$代入:
$\frac {\frac {3}{2}+5}{\frac {3}{2}}=\frac {\frac {13}{2}}{\frac {3}{2}}=\frac {13}{3}$