[tex]\displaystyle\bf\\x\in Z\\\\\frac{-5}{3x-1}\in Z\\\\\implies~(3x-1)\in D_{-5}\\\\D_{-5}=\{-5;~-1;~1;~5\}\\\\3x-1=-5\\3x=-5+1\\3x=-4\\x=-\frac{4}{3}~\notin Z~~(Solutie~eliminata)\\\\\\3x-1=-1\\3x=-1+1\\3x=0\\x=0\in Z\\\\\\3x-1=1\\3x=1+1\\3x=2\\x=\frac{2}{3}~\notin Z~~(Solutie~eliminata)\\\\\\3x-1=5\\3x=5+1\\3x=6\\\\x=\frac{6}{3}\\\\x=2\in Z\\\\\\solutiile~sunt:\\\boxed{\bf x_1=0~~si~~x_2=2}\\\boxed{\bf x\in\{0;~2\}}[/tex]
