\(B=\frac{2-x}{x+3}=\frac{-\left(x+3\right)+1}{x+3}=-1+\frac{1}{x+3}\)
\(ĐểB\in Z\) với \(x\in Z\) thì\(\frac{1}{x+3}\in Z\Leftrightarrow1\)chia hết cho (x+3)
\(\Leftrightarrow\left(x+3\right)\inƯ\left(1\right)\)
Mà \(Ư\left(1\right)\in\left\{-1;1\right\}\)
th1: \(x+3=1\Leftrightarrow x=-2\)
TH2:\(x+3=-1\Leftrightarrow x=-4\)