Answer
$\frac{t-3}{3}$
Work Step by Step
$\frac{t^{2}-9}{3t+9}=\frac{(t+3)(t-3)}{3(t+3)}=\frac{t-3}{3}\times\frac{t+3}{t+3}=\frac{t-3}{3}\times1=\frac{t-3}{3}$
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