D = {-6; -2; 0; 4}
Vezi rezolvarea mai jos.
.
[tex]\displaystyle\bf\\D=\left\{x\in Z\left|~\frac{2x-3}{x+1}\in Z\right\}\right\\\\-3=2-5\\\\\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2x+2}{x+1}+\frac{-5}{x+1}=\\\\\frac{2(x+1)}{x+1}-\frac{5}{x+1}=2-\frac{5}{x+1}\\\\2\in Z\\\\\implies~\frac{5}{x+1}\in Z\\\\\implies~(x+1)\in D_5\\\\D_5=\{-5;~-1;~1;~5\}\\\\x+1=-5 \implies x=-6\\x+1=-1 \implies x=-2\\x+1=1 \implies x=0\\x+1=5 \implies x=4\\\\\boxed{\bf D=\{-6;~-2;~0;~4\}}[/tex]
