\(a,\Rightarrow n+2+3⋮n+2\\ \Rightarrow n+2\inƯ\left(3\right)=\left\{1;3\right\}\\ \Rightarrow n=1\left(n\in N\right)\\ b,\Rightarrow n-2+7⋮n-2\\ \Rightarrow n-2\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n=5\left(n\in N\right)\\ c,\Rightarrow\left(n^2-n\right)+\left(3n-3\right)+3⋮n-1\\ \Rightarrow n\left(n-1\right)+3\left(n-1\right)+3⋮n-1\\ \Rightarrow n-1\inƯ\left(3\right)=\left\{1;3\right\}\\ \Rightarrow n\in\left\{2;4\right\}\)
a: \(\Leftrightarrow n+2=3\)
hay n=1