ĐKXĐ: \(n\ne-\dfrac{1}{2}\)
Để A nguyên thì \(3n+2⋮2n+1\)
\(\Leftrightarrow2\left(3n+2\right)⋮2n+1\)
\(\Leftrightarrow6n+4⋮2n+1\)
\(\Leftrightarrow6n+3+1⋮2n+1\)
mà \(6n+3⋮2n+1\)
nên \(1⋮2n+1\)
\(\Leftrightarrow2n+1\inƯ\left(1\right)\)
\(\Leftrightarrow2n+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow2n\in\left\{0;-2\right\}\)
hay \(n\in\left\{0;-1\right\}\)(thỏa mãn)
Vậy: Để A nguyên thì \(n\in\left\{0;-1\right\}\)