a: \(B=\dfrac{18}{\left(x-3\right)^2\cdot\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)^2\cdot\left(x+3\right)}-\dfrac{x^2-3x}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{9-x^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-1}{x-3}\)
\(a,B=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-1}{x-3}\\ b,\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ \text{Với }x=2\Leftrightarrow B=\dfrac{-1}{2-3}=-1\\ \text{Với }x=-5\Leftrightarrow B=\dfrac{-1}{-5-3}=\dfrac{1}{8}\\ c,B\in Z\\ \Leftrightarrow x-3\inƯ\left(-1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-2;4\right\}\)
