\(\frac{2n+1}{n-1}=\frac{2n-2+3}{n-1}=\frac{2n-2}{n-1}+\frac{3}{n-1}\)\(=\frac{2.\left(n-1\right)}{n-1}+\frac{3}{n-1}=2+\frac{3}{n-1}\)
\(\Rightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n-1=1\Rightarrow n=2\)
\(n-1=-1\Rightarrow n=0\)
\(n-1=3\Rightarrow n=4\)
\(n-1=-3\Rightarrow n=-2\)
Vậy \(x\in\left\{\pm2;0;4\right\}\)
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