Đáp án:
$\begin{array}{l}
A = \left( {\frac{{x - 3}}{x} - \frac{x}{{x - 3}} + \frac{9}{{{x^2} - 3x}}} \right):\frac{{2x - 2}}{x}\\
Đkxđ:x \ne 0;x \ne 3;x \ne 1\\
A = \frac{{\left( {x - 3} \right)\left( {x - 3} \right) - x.x + 9}}{{x\left( {x - 3} \right)}}.\frac{x}{{2x - 2}}\\
= \frac{{{x^2} - 6x + 9 - {x^2} + 9}}{{\left( {x - 3} \right)\left( {2x - 2} \right)}}\\
= \frac{{18 - 6x}}{{\left( {x - 3} \right)\left( {2x - 2} \right)}}\\
= \frac{{ - 3\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x - 1} \right)}}\\
= \frac{{ - 3}}{{x - 1}}\\
b)x \ne 0;x \ne 3;x \ne 1\\
A = 2\\
\Rightarrow - \frac{3}{{x - 1}} = 2\\
\Rightarrow 2x - 2 = - 3\\
\Rightarrow x = - \frac{1}{2}\left( {tmdk} \right)\\
c)A = \frac{{ - 3}}{{x - 1}} \in Z\\
\Rightarrow \left( {x - 1} \right) \in Ư\left( 3 \right) = {\rm{\{ }} - 3; - 1;1;3\} \\
\Rightarrow x \in {\rm{\{ }} - 2;0;2;4\} \\
x \ne 0 \Rightarrow x \in {\rm{\{ }} - 2;2;4\}
\end{array}$
