a/ Ta có :
\(A=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{12n+1}{2n+3}=\dfrac{1}{3}\)
\(\Leftrightarrow3\left(12n+1\right)=2n+3\)
\(\Leftrightarrow36n+3=2n+3\)
\(\Leftrightarrow36n-2n=3-3\)
\(\Leftrightarrow34n=0\)
\(\Leftrightarrow n=0\)
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b/ Để Phân số \(A\in Z\) thì :
\(12n+1⋮2n+3\)
Mà \(2n+3⋮2n+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}12n+1⋮2n+3\\6n+18⋮2n+3\end{matrix}\right.\)
\(\Leftrightarrow17⋮2n+3\)
\(\Leftrightarrow2n+3\inƯ\left(17\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2n+3=1\\2n+3=17\\2n+3=-1\\2n+3=-17\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-1\left(loại\right)\\n=\dfrac{15}{2}\\n=-2\left(loại\right)\\-\dfrac{21}{2}\left(loại\right)\end{matrix}\right.\)
Vậy ..
