Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 0;x \ne 3;x \ne - 3;x \ne - \dfrac{3}{2}\\
P = \left( {\dfrac{{x + 3}}{{{x^2} - 3x}} - \dfrac{x}{{{x^2} - 9}}} \right):\dfrac{{2x + 3}}{{{x^2} + 3x}}\\
= \left( {\dfrac{{x + 3}}{{x\left( {x - 3} \right)}} - \dfrac{x}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right).\dfrac{{x\left( {x + 3} \right)}}{{2x + 3}}\\
= \dfrac{{{{\left( {x + 3} \right)}^2} - x.x}}{{x\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x\left( {x + 3} \right)}}{{2x + 3}}\\
= \dfrac{{{x^2} + 6x + 9 - {x^2}}}{{x - 3}}.\dfrac{1}{{2x + 3}}\\
= \dfrac{{6x + 9}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{{3\left( {2x + 3} \right)}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}\\
= \dfrac{3}{{x - 3}}\\
c)P = \dfrac{3}{{x - 3}} \in Z\\
\Leftrightarrow \left( {x - 3} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ {0;2;4;6} \right\}\\
Do:x \ne 0;x \ne 3;x \ne - 3\\
\Leftrightarrow x \in \left\{ {2;4;6} \right\}\\
Vay\,x \in \left\{ {2;4;6} \right\}
\end{array}$
