[tex]\displaystyle\\Cel~mai~mic~numar~intreg~x~pentru~care~\frac{x-1}{x+1}~apartine~lui~\mathbb{Z}~este~x=-3,\\deoarece~~\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=\frac{x+1}{x+1}-\frac{2}{x+1}=1-\frac{2}{x+1}~si~cum~1\in\mathbb{Z} \Longrightarrow\\\frac{2}{x+1}\in\mathbb{Z},~de~unde~(x+1)|2 \Longleftrightarrow (x+1)\in\mathcal{D}_2 \Longleftrightarrow x+1\in\left\{-2,-1,1,2\right\},\\de~unde~x\in\left\{-3,-2,0,1\right\},~cum~problema~ne~cere~minimul~lui~x~obtinem\\x=-3.[/tex]
