a,
để A là một phân số thì \(2\times n+3\ne0\)và \(n\ne-\frac{3}{2}\)
b,
\(\frac{12\times n+1}{2\times n+3}=\frac{6\times\left(2\times n+3\right)-18+1}{2\times n+3}=\)\(=\frac{6\times\left(2\times n+3\right)}{2\times n+3}+\frac{-17}{2\times n+3}\)\(=6+-\frac{17}{2\times n+3}\)
\(A\in Z\Rightarrow-\frac{17}{2\times n+3}\in Z\Rightarrow\)\(2\times n+3\in U\left(-17\right)=\hept{ }1;-1;17;-17\)
\(\Rightarrow n\in\left(-1;-2;7;-10\right)\)thì A nguyên