Question

dau coroana! ajutor plssss

Answer 1

$\frac{3}{2n+1}\in N~\Leftrightarrow~(2n+1)\in~D_3.\\ \\ 2n+1=1~cu~n=0\\ 2n+1=-1~cu~n=-1\\ 2n+1=3~cu~n=1\\ 2n+1=-3~cu~n=-2\\ \\ Deoarece~n\in~N,~avem~solutiile~\{0,1\}.$