5n\(⋮\)n-2
5n-10+10\(⋮\)n-2
5(n-2)+10\(⋮\)n-2
Vì 5(n-2)\(⋮\)n-2
Buộc 10\(⋮\)n-2=>n-2 ϵ Ư(10)={1;2;5;10}
ta có bảng sau :
| n-2 | 1 | 2 | 5 | 10 |
| n | 3 | 4 | 7 | 8 |
vậy n ϵ {3;4;7;8}
3n+4\(⋮\)n+1
3n+3+1\(⋮\)n+1
3(n+1)+1\(⋮\)n+1
Vì 3(n+1)\(⋮\)n+1
Buộc 1 \(⋮\)n+1=>n+1ϵƯ(1)={1}
Với n+1=1=>n=0
Vậy n ϵ {0}
5n\(⋮\)n+2
5n+10-10\(⋮\)n+2
5(n+2)-10\(⋮\)n+2
Vì5(n+2)\(⋮\)n+2
Buộc 10 \(⋮\)n+2=>n+2ϵƯ(10)={1;2;5;10}
ta có bảng sau :
| n+2 | 1 | 2 | 5 | 10 |
| n | -1 | 0 | 3 | 8 |
Vậy n ϵ {0;3;8}
\(5n⋮n-2\)
\(\Leftrightarrow5n-10+10⋮n-2\)
\(\Leftrightarrow5\left(n-2\right)+10⋮n-2\)
\(\Leftrightarrow10⋮n-2\)
\(\Leftrightarrow n-2\in\text{Ư}\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Leftrightarrow n\in\left\{-8;-3;0;1;3;4;7;12\right\}\)
mà \(n\in N\)
\(\Rightarrow n\in\left\{0;1;3;4;7;12\right\}\)
\(3n+4⋮n+1\)
\(\Leftrightarrow3n+3+1⋮n+1\)
\(\Leftrightarrow3\left(n+1\right)+1⋮n+1\)
\(\Leftrightarrow1⋮n+1\)
\(\Leftrightarrow n+1\in\text{Ư}\left(1\right)=\left\{-1;1\right\}\)
\(\Leftrightarrow n\in\left\{-2;0\right\}\)
mà \(n\in N\)
=> n = 0
\(5n⋮n+2\)
\(\Leftrightarrow5n+10-10⋮n+2\)
\(\Leftrightarrow5\left(n+2\right)-10⋮n+2\)
\(\Leftrightarrow-10⋮n+2\)
\(\Leftrightarrow n+2\in\text{Ư}\left(-10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Leftrightarrow n\in\left\{-12;-7;-4;-3;-1;0;3;8\right\}\)
mà \(n\in N\)
\(\Rightarrow n\in\left\{0;3;8\right\}\)
giúp mk nha
