Ta có :
$A = \dfrac{x^2-6x+9}{2x^2-6x} $
a) $ĐKXĐ : \left\{ \begin{array}{l}x^2-6x+9 \neq 0\\2x^2-6x \neq 0\end{array} \right.$
$⇔ \left\{ \begin{array}{l}(x-3)^2 \neq 0\\2x(x-3) \neq 0\end{array} \right.$
$⇔ \left\{ \begin{array}{l}x \neq 0\\x \neq 3\end{array} \right.$
b) $A = \dfrac{x^2-6x+9}{2x^2-6x} $
$=\dfrac{(x-3)^2}{2x.(x-3)}$
$=\dfrac{x-3}{2x} $
c) Để $A=-1$
$⇔\dfrac{x-3}{2x} = -1$
$⇔ \dfrac{x-3+2x}{2x} = 0$
$⇒3x-3=0$
$⇔x=1$ ( Thỏa mãn )