Ta có : \(\frac{3n+1}{2n-3}\Rightarrow\frac{2\left(3n+1\right)}{3\left(2n-3\right)}\Rightarrow\frac{6n+2}{6n-9}\)
\(\Rightarrow\left(6n+2\right)-\left(6n-9\right)⋮2n-3\)
\(\Rightarrow6n+2-6n+9⋮2n-3\)
\(\Rightarrow11⋮2n-3\Rightarrow2n-3\inƯ\left(11\right)\)
Mà : \(Ư\left(11\right)=\left\{1;11\right\}\Rightarrow2n-3\in\left\{1;11\right\}\)
+) \(2n-3=1\Rightarrow2n=1+3\Rightarrow2n=4\Rightarrow n=2\)
+) \(2n-3=11\Rightarrow2n=11+3\Rightarrow2n=14\Rightarrow n=7\)
Vậy : \(n\in\left\{2;7\right\}\)
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