Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
{x^2} - 3x \ne 0\\
3 - x \ne 0\\
x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 3\\
x \ne 0
\end{array} \right.\\
b)A = \dfrac{{ - 15}}{{{x^2} - 3x}} - \dfrac{5}{{3 - x}} - \dfrac{2}{x}\\
= \dfrac{{15}}{{x\left( {3 - x} \right)}} - \dfrac{5}{{3 - x}} - \dfrac{2}{x}\\
= \dfrac{{15 - 5.x - 2\left( {3 - x} \right)}}{{x\left( {3 - x} \right)}}\\
= \dfrac{{15 - 5x - 6 + 2x}}{{x\left( {3 - x} \right)}}\\
= \dfrac{{9 - 3x}}{{x\left( {3 - x} \right)}}\\
= \dfrac{3}{x}\\
c)A \in N*\\
\Rightarrow 3 \vdots x\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = 3\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 1
\end{array}$
