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