\(\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}\)\(=\frac{2\left(x-3\right)}{x-3}+\frac{7}{x-3}=2+\frac{7}{x-3}\)
\(\Rightarrow\)\(x-3\inƯ\left(7\right)=\left\{1;7\right\}\)
\(\Rightarrow x-3=1\Rightarrow x=4\)
\(x-3=7\Rightarrow x=10\)
Vậy \(x\in\left\{4;10;2\right\}\)
NHỚ TK NHA