\(3n-1⋮n+2\)
\(\Rightarrow3n+6-7⋮n+2\)
\(\Rightarrow3\left(n+2\right)-7⋮n+2\)
Do : \(3\left(n+2\right)⋮n+2\)suy ra : \(7⋮n+2\)
\(\Rightarrow n+2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow n\in\left\{-9;-3;-1;5\right\}\)
3n - 1 = 3n + 6 - 7 = 3(n + 2) - 7
Để \(\left(3n-1\right)⋮\left(n+2\right)\) thì \(7⋮\left(n+2\right)\)
=> \(n+2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Nếu n + 2 = -7 => n = -7 - 2 => n = -9
Nếu n + 2 = -1 => n = -1 - 2 => n = -3
Nếu n + 2 = 1 => n = 1 - 2 => n = -1
Nếu n + 2 = 7 => n = 7 - 2 => n = 5
Vậy \(n\in\left\{-9;-3;-1;5\right\}\)
a)( 3n - 1 ) \(⋮\)n + 2 \(\Rightarrow\) 3(n+2)\(⋮\)n+2 => 3n+4 \(⋮\) n+2
=> 3n+4 - 3n-1 \(⋮\)n+2
3\(⋮\)n+2
n+2 | 1 | 3 |
n | -1 | 1 |
vậy n = -1 : 1
b) 5n+3\(⋮\)2n+1 => 2(5n+3) - 5(2n+1)\(⋮\)2n+1
10n+3-10n+5 \(⋮\)2n+1
2 chia hết 2n+1
2n+1 | 1 | 2 |
n | 0 | loại |
vậy n = 0