ĐKXĐ: \(2n+1\ne0\Leftrightarrow n\ne-0,5\)
Với \(n\ne0,5\) thì \(A=\dfrac{12n+4}{2n+1}\)
\(=\dfrac{12n+6-2}{2n+1}\)
\(=\dfrac{6\left(2n+1\right)}{2n+1}-\dfrac{2}{2n+1}\)
\(=6-\dfrac{2}{n+1}\)
Để \(A\in Z\) thì \(2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)\)
\(\Rightarrow n+1\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow n\in\left\{-3;-2;0;1\right\}\) (tmđk)
Vậy để \(A\in Z\) thì \(n\in\left\{-3;-2;0;1\right\}\)
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