\(4n-7⋮6n+1\)
\(\Rightarrow3\left(4n-7\right)⋮6n+1\)
\(\Rightarrow12n-21⋮6n+1\)
\(\Rightarrow\left(12n+2\right)-23⋮6n+1\)
\(\Rightarrow2\left(6n+1\right)-23⋮6n+1\)
\(\Rightarrow23⋮6n+1\)
\(\Rightarrow6n+1\in\left\{1;-1;23;-23\right\}\)
\(\left[\begin{matrix}6n+1=1\\6n+1=-1\\6n+1=23\\6n+1=-23\end{matrix}\right.\Rightarrow\left[\begin{matrix}n=0\\n=-\frac{1}{3}\\n=\frac{11}{3}\\n=-4\end{matrix}\right.\)
Vậy \(n\in\left\{0;\frac{-1}{3};\frac{11}{3};-4\right\}\)
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