Ta có : \(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)
Để \(\left(n+1\right)⋮\left(n-2\right)\)thì \(3⋮\left(n-2\right)\)hay \(\left(n-2\right)\)là \(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
Do đó :
n - 2 | 1 | -1 | 3 | -3 |
n | 3 | 1 | 5 | -1 |
Vậy .....................
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