[tex]\it B=\{x\in\mathbb{Z}\Big |\ \dfrac{18}{x} \in\mathbb{Z},\ |x|=-x\}\\ \\ \\ |x|=-x\ \Rightarrow x<0\ \ \ \ \ \ (1)\\ \\ \\ \dfrac{18}{x}\in\mathbb{Z} \Rightarrow x\in\ D_{18}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow x\in\{-18,\ -9,\ -6,\ -3,\ -2,\ -1\}=B[/tex]
