$\dfrac{x^2-3x+5}{(x-3)(x+2)}=\dfrac{1}{x-3}\hspace{1cm}(x\neq3;x\neq-2)$
$⇒x^2-3x+5=x+2$
$⇔x^2-3x+5-x-2=0$
$⇔x^2-4x+3=0$
$⇔x^2-x-3x+3=0$
$⇔x(x-1)-3(x-1)=0$
$⇔(x-1)(x-3)=0$
\(⇔\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1(tmdk)\\x=3(ktmdk)\end{array} \right.\)
Vậy phương trình có nghiệm duy nhất là $x=1$.