Ta có : \(A=\dfrac{12n+1}{2n+3}=\dfrac{6\left(2n+3\right)-17}{2n+3}=6-\dfrac{17}{2n+3}\)
\(\Rightarrow2n+3\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
| 2n + 3 | 1 | -1 | 17 | -17 |
| 2n | -2 | -4 | 14 | -20 |
| n | -1 | -2 | 7 | -10 |
Để A là số nguyên thì 12n+1⋮2n+3
12n+18-17⋮2n+3
12n+18⋮2n+3 ⇒17⋮2n+3
2n+3∈Ư(17)
Ư(17)={1;-1;17;-17}
n∈{-1;-2;7;-10}
Để A nguyên thì \(12n+1⋮2n+3\)
\(\Leftrightarrow-17⋮2n+3\)
\(\Leftrightarrow2n+3\inƯ\left(-17\right)\)
\(\Leftrightarrow2n+3\in\left\{1;-1;17;-17\right\}\)
\(\Leftrightarrow2n\in\left\{-2;-4;14-20\right\}\)
\(\Leftrightarrow n\in\left\{-1;-2;7;-10\right\}\)
