Đặt \(A=\frac{6n-3}{3n+1}=\frac{\left(6n+2\right)-2-3}{3n+1}=\frac{2.\left(3n+1\right)-5}{3n+1}\)
\(\Rightarrow A=\frac{2.\left(3n+1\right)}{3n+1}-\frac{5}{3n+1}=2-\frac{5}{3n+1}\)
\(A\in Z\Leftrightarrow\frac{5}{3n+1}\in Z\Leftrightarrow5⋮\left(3n+1\right)\Leftrightarrow\left(3n+1\right)\inƯ\left(5\right)\)
=> 3n + 1 \(\in\){1;-1;5;-5}
Ta có bảng :
| 3n+1 | 1 | -1 | 5 | -5 |
| n | 0 | \(-\frac{2}{3}\) | \(\frac{4}{3}\) | -2 |
Mà \(n\in Z\)=>\(n\in\){0;-2} để phân số \(\frac{6n-3}{3n+1}\in Z\)
để \(\frac{6n-3}{3n+1}\)là số nguyên thì 6n-3\(⋮\)3n-1
ta có \(\orbr{\begin{cases}6n-3⋮3n+1\\3n+1⋮3n+1\end{cases}}\Rightarrow\orbr{\begin{cases}6n-3⋮3n+1\\2\left(3n+1\right)⋮3n+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}6n-3⋮3n+1\\6n+2⋮3n+1\end{cases}}\)
\(\Rightarrow\left(6n+2\right)-\left(6n-3\right)\)\(⋮3n+1\)
\(5⋮3n+1\)
=>3n+1\(\in\)Ư(5)={-1,-5,1,5}
ta co bang sau
...
Ta có: \(\frac{6n-3}{3n+1}\)=\(\frac{2\left(3n+1\right)-5}{3n+1}\)= 2-\(\frac{5}{3n-1}\)
Từ đó để \(\frac{6n-3}{3n+1}\) \(\varepsilon\)Z thì 3n+1 \(\varepsilon\)Ư(5)
Ta có bảng sau
| 3n+1 | -5 | -1 | 1 | 5 |
| n | -2 | \(\frac{-2}{3}\)(Loại) | 0 | \(\frac{4}{3}\)(Loại) |
Vậy với n\(\varepsilon\){-2;0} thì \(\frac{6n-3}{3n+1}\varepsilon\)Z
