#### Answer

$\dfrac{x-3}{x^{2}-6x+9}=\dfrac{1}{x-3}$

#### Work Step by Step

$\dfrac{x-3}{x^{2}-6x+9}$
The denominator is a perfect square trinomial. Factor it and then simplify:
$\dfrac{x-3}{x^{2}-6x+9}=\dfrac{x-3}{(x-3)^{2}}=\dfrac{1}{x-3}$

Published by
Pearson

ISBN 10:
0321726391

ISBN 13:
978-0-32172-639-1

$\dfrac{x-3}{x^{2}-6x+9}=\dfrac{1}{x-3}$

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