các bạn ơi 12n+5 nhé mình viết thiếu mất :)
Cho \(A=\frac{12n+5}{2n+3}=\frac{6\left(2n+3\right)-13}{2n+3}=\frac{6\left(2n+3\right)}{2n+3}-\frac{13}{2n+3}\in Z\)
Để \(A\in Z\Rightarrow13⋮\left(2n+3\right)\)hay \(2n+3\inƯ\left(13\right)\)
Ta có :
\(Ư\left(13\right)\in\left\{\pm1;\pm13\right\}\Rightarrow2n+3\in\left\{\pm1;\pm13\right\}\)
| \(2n+3\) | \(n\) |
| \(1\) | \(-1\) |
| \(-1\) | \(-2\) |
| \(13\) | \(5\) |
| \(-13\) | \(-8\) |
Vậy để A nguyên \(\Rightarrow n\in\left\{-1;-2;5;-8\right\}\)
