a: Để A nguyên thì \(n+2=11\)
hay n=9
b: Để B nguyên thì \(n-3\in\left\{-1;1;11\right\}\)
hay \(n\in\left\{2;4;14\right\}\)
\(a,\Rightarrow\dfrac{5\left(n-1\right)+4}{n+2}\in Z\Rightarrow n+2\inƯ\left(4\right)=\left\{1;2;4\right\}\\ \Rightarrow n\in\left\{0;2\right\}\left(n\in N\right)\\ b,\Rightarrow\dfrac{3\left(n-3\right)+11}{n-3}\in Z\\ \Rightarrow n-3\inƯ\left(11\right)=\left\{1;11\right\}\\ \Rightarrow n\in\left\{4;14\right\}\\ c,\Rightarrow\dfrac{2\left(n+2\right)-5}{n+2}\in Z\\ \Rightarrow n+2\inƯ\left(5\right)=\left\{1;5\right\}\\ \Rightarrow n=3\left(n\in N\right)\)
