ta xét:
\(\frac{5n+2}{9-2n}=\frac{9-2n-7+7n}{9-2n}=\frac{9-2n}{9-2n}-\frac{7-7n}{9-2n}\)
\(=1-7.\frac{\left(1-n\right)}{9-2n}\)
Để \(\left(5n+2\right)⋮\left(9-2n\right)\Leftrightarrow\left(1-n\right)⋮\left(9-2n\right)\)
Giả sử \(\left(1-n\right)⋮\left(9-2n\right)\)
\(\Rightarrow2\left(1-n\right)⋮\left(9-2n\right)\)
\(\Rightarrow\left(2-2n\right)⋮\left(9-2n\right)\)
\(\left(9-2n-7\right)⋮\left(9-2n\right)\)
Vì \(\left(9-2n\right)⋮\left(9-2n\right)\Rightarrow-7⋮\left(9-2n\right)\)
\(\Leftrightarrow9-2n\inƯ\left(7\right)\Rightarrow9-2n\in\left(\pm1;\pm7\right)\)
ta có bảng sau
| 9-2n | -7 | -1 | 1 | 7 |
| 2n | 16 | 10 | 8 | 2 |
| n | 8 (loại vì n<5) | 5 (loại vì n<5 > | 4 | 1 |
Vậy để \(\left(5n+2\right)⋮\left(9-2n\right)\Leftrightarrow n\in\left(1;4\right)\)
