Để \(A\in Z\Rightarrow4n-1⋮2n+3\)
Ta có:\(\left\{{}\begin{matrix}4n-1⋮2n+3\\2n+3⋮2n+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4n-1⋮2n+3\\2\left(2n+3\right)⋮2n+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4n-1⋮2n+3\\4n+6⋮n+3\end{matrix}\right.\)
\(\Rightarrow\left(4n+6\right)-\left(4n-1\right)⋮2n+3\)
\(\Rightarrow4n+6-4n+1⋮2n+3\Rightarrow4n-4n+6-1⋮2n+3\)
\(\Rightarrow5⋮2n+3\Rightarrow2n+3\inƯ\left(5\right)\)
\(\Rightarrow2n+3\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow2n\in\left\{-8;-4;\pm2\right\}\)
\(\Rightarrow n\in\left\{-4;-2;\pm1\right\}\)
Vậy \(n\in\left\{-4;-2;\pm1\right\}\)
Để A là số nguyên thì 4n - 1 \(⋮\) 2n + 3
Có: 2n + 3 \(⋮\) 2n + 3
=> 2(2n + 3) \(⋮\) 2n + 3
=> 4n + 6 \(⋮\) 2n + 3
Xét hiệu:
(4n + 6) - (4n - 1) \(⋮\) 2n + 3
=> 4n + 6 - 4n + 1 \(⋮\) 2n + 3
=> 7 \(⋮\) 2n + 3
Vì 4n - 1 \(⋮\) 2n + 3 nên 2n + 3 \(\in\) Ư(7) = \(\left\{-7;-1;1;7\right\}\)
+) 2n + 3 = -7
2n = -10
=> n = -5
+) 2n + 3 = -1
2n = -4
=> n = -2
+) 2n + 3 = 1
2n = -2
=> n = -1
+) 2n + 3 = 7
2n = 4
=> n = 2
Vậy n \(\in\)\(\left\{-5;-2;-1;2\right\}\)
