\(\dfrac{6n+5}{2n-1}=\dfrac{6n-3+8}{2n-1}=\dfrac{3\left(2n-1\right)+8}{2n-1}=\dfrac{3\left(2n-1\right)}{2n-1}+\dfrac{8}{2n-1}\)
\(\Rightarrow8⋮2n-1\Rightarrow2n-1\inƯ\left(8\right)\)
\(Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
2n-1 lẻ:
\(2n-1=1\Rightarrow2n=2\Rightarrow n=1\)
\(2n-1=-1\Rightarrow2n=0\Rightarrow n=0\)
=> 6n+5 chia hết cho 2n-1
=> 6n-3+8 chia hết cho 2n-1
=> 3(2n-1)+8 chia hết cho 2n-1
=> 8 chia hết cho 2n-1
=> \(2n-1=-1;1;-2;2;-4;4;-8;8\)
Vì 2n-1 là số lẽ
=> \(\text{2n-1=-1;1}\)
=> \(\text{2n=0;2}\)
=> \(\text{n=0;1}\)