https://www.google.com/search?q=3n%2B2+chia+h%E1%BA%BFt+cho+2n-1&oq=3n%2B2+chia+h%E1%BA%BFt+cho+2n-1&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8
\(3n+2⋮2n-1;2\in N\)
\(\Rightarrow2.\left(3n+2\right)⋮2n-1\)
\(\Rightarrow6n+4⋮2n-1\)
\(2n-1⋮2n-1;3\in N\)
\(\Rightarrow3.\left(2n-1\right)⋮2n-1\)
\(\Rightarrow6n-3⋮2n-1\)
Do vậy:
\(\left(6n+4\right)-\left(6n-3\right)⋮2n-1\)
\(\Rightarrow7⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(7\right)\)
\(\Rightarrow2n-1\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow n\in\left\{1;0;4;-3\right\}\)
Ta có: \(3n+2⋮2n-1\)
\(\Leftrightarrow2\cdot\left(3n+2\right)⋮2n-1\)
\(\Leftrightarrow6n+4⋮2n-1\)
\(\Leftrightarrow6n-3+7⋮2n-1\)
mà \(6n-3⋮2n-1\)
nên \(7⋮2n-1\)
\(\Leftrightarrow2n-1\inƯ\left(7\right)\)
\(\Leftrightarrow2n-1\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2n\in\left\{2;0;8;-6\right\}\)
hay \(n\in\left\{1;0;4;-3\right\}\)
Vậy: \(n\in\left\{1;0;4;-3\right\}\)
