Ta có : C = \(\frac{9n-2}{3n+1}=\frac{9n+3-5}{3n+1}=\frac{3\left(3n+1\right)-5}{3n+1}=3-\frac{5}{3n+1}\)
Vì \(3\inℤ\)
=> \(C\inℤ\Leftrightarrow\frac{-5}{3n+1}\inℤ\Rightarrow-5⋮3n+1\Rightarrow3n+1\inƯ\left(-5\right)\)
=> \(3n+1\in\left\{1;5;-1;-5\right\}\)
=> \(3n\in\left\{0;4;-2;-6\right\}\)
Vì n \(\inℤ\)
=> \(n\in\left\{0;-2\right\}\)
Bg
Để \(C=\frac{9n-2}{3n+1}\inℤ\)(n \(\inℕ\)) thì 9n - 2 \(⋮\)3n + 1
Vì 9n - 2 \(⋮\)3n + 1
Nên (9n - 2) - 3.(3n + 1) \(⋮\)3n + 1
=> 9n - 2 - 9n + 9 \(⋮\)3n + 1
=> 9n - 9n + (9 - 2) \(⋮\)3n + 1
=> 7 \(⋮\)3n + 1
=> 3n + 1 \(\in\)Ư(7)
Ư(7) = {1; 7}
=> 3n + 1 = 1 hay 7
3n = 1 - 1 hay 7 - 1
3n = 0 hay 6
n = 0 : 3 hay 6 : 3
n = 0 hay 2
Vậy n = 0 hoặc n = 2
\(C=\frac{9n-2}{3n+1}=\frac{3\left(3n+1\right)-5}{3n+1}=3-\frac{5}{3n+1}\)
Để C nguyên => \(\frac{5}{3n+1}\)nguyên
=> \(5⋮3n+1\)
=> \(3n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
| 3n+1 | 1 | -1 | 5 | -5 |
| n | 0 | -2/3 | 4/3 | -2 |
Chưa rõ đkxđ của n nên để tạm vầy nhé !
\(C=\frac{9n-2}{3n+1}=\frac{3\left(3n+1\right)-5}{3n+1}=\frac{-5}{3n+1}\)
\(\Rightarrow3n+1\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
| 3n + 1 | 1 | -1 | 5 | -5 |
| 3n | 0 | -2 | 4 | -6 |
| n | 0 | -2/3 | 4/3 | -2 |
Vì \(x\in Z\Rightarrow x=0;-2\)