Để \(C=\dfrac{3n+5}{2n+3}\) là 1 số nguyên thì :
\(3n+5⋮2n+3\)
Mà \(2n+3⋮2n+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}6n+10⋮2n+3\\6n+9⋮2n+3\end{matrix}\right.\)
\(\Leftrightarrow1⋮2n+3\)
\(\Leftrightarrow2n+3\inƯ\left(1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2n+3=1\\2n+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=-1\\n=-2\end{matrix}\right.\)
Vậy ,,
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