Ta có : \(n+4=n-1+\)\(5\)
Ta thấy : \(\left(n-1\right)⋮\left(n-1\right)\)
Nên \(\left(n+4\right)⋮\left(n-1\right)\Leftrightarrow5⋮\)\(\left(n-1\right)\)
\(\Leftrightarrow\left(n-1\right)\inƯ\left(5\right)=\)\((1;5)\)
| N - 1 | 1 | 5 |
| N | 2 | 6 |
a) \(n+4⋮n-1\Rightarrow\left(n-1\right)+5⋮n-1\Rightarrow5⋮n-1\Rightarrow n-1\inƯ\left(5\right)\)
\(\Rightarrow n-1\in\left\{1;5;-1;-5\right\}\Rightarrow n\in\left\{2;6;0;-4\right\}\)
b) \(n^2+2n-3=\left(n^2+n\right)+n-3=n\left(n+1\right)+n-3\)
vì \(n\left(n-1\right)⋮n-1\)\(\Rightarrow n-3⋮n+1\Rightarrow\left(n+1\right)-4⋮n-1\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)
\(\Rightarrow n-1\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow n\in\left\{2;3;5;0;-1;-3\right\}\)
\(\left(n^2+2n-3\right)⋮\left(n+1\right)\)
\(n^2+2n-3=\)\(n^2+n+n-3\)
\(=n.\left(n+1\right)+n+1-4\)
Mà \(n.\left(n+1\right)⋮\left(n+1\right)\)
\(\left(n+1\right)⋮\left(n+1\right)\)
Nên \(n^2+2n-3⋮\left(n+1\right)\) \(\Leftrightarrow4⋮\left(n+1\right)\)
\(\Leftrightarrow\left(n+1\right)\inƯ\left(4\right)=\)\((1;2;4)\)
| n+1 | 1 | 2 | 4 |
| n | 0 | 1 | 3 |
