a) \(A=\dfrac{6n-1}{3n+2}=\dfrac{6n+4-5}{3n+2}=\dfrac{2\left(3n+2\right)-5}{3n+2}=2-\dfrac{5}{3n+2}\)
Để A nguyên thì \(5⋮3n+2\) \(\Rightarrow3n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow n\in\left\{-1;1\right\}\)
b) \(B=\dfrac{3n-5}{n+4}=\dfrac{3n+12-7}{n+4}=\dfrac{3\left(n+4\right)-7}{n+4}=3-\dfrac{7}{n+4}\)
Để B nguyên thì \(7⋮n+4\) \(\Rightarrow n+4\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow n\in\left\{-11;-5;-3;3\right\}\)