a,vì (n+2) ⋮ (n-1)
(n-1)⋮(n-1)
=>(n+2)-(n-1)⋮(n-1)
=>n+2-n+1⋮(n-1)
=>3⋮(n-1)
=>(n-1)∈Ư(3)={1;3}
ta có bảng
| n-1 | 1 | 3 |
| n | 2 | 4 |
| ✔ | ✔ |
vậy n=2 hoặc4
d, (2n+1) ⋮ (6-n)
vì (6-n)⋮(6-n)
=> 2(6-n)⋮(6-n)
=> (12-2n)⋮(6-n)
=> (2n+1)+(12-2n)⋮(6-n)
=> (2n+1+12-2n)⋮(6-n)
=> 13⋮(6-n)
=> 6-n ∈Ư(13)={-13;-1;1;13}
ta có bảng sau
| 6-n | -13 | -1 | 1 | 13 |
| n | 19 | 7 | 5 | -7 |
| nx | t/m | t/m | tm | loại |
vậy n ∈{5;7;19}
a) Ta có:
\(n+2⋮n-1\)
\(\Rightarrow\left(n-1\right)+3⋮n-1\)
\(\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\in\left\{1;3\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=3\Rightarrow n=4\end{matrix}\right.\)
Vậy n=2 hoặc n=4
b) Ta có:
\(2n+7⋮n+1\)
\(\Rightarrow\left(2n+2\right)+5⋮n+1\)
\(\Rightarrow2\left(n+1\right)+5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=5\Rightarrow n=4\end{matrix}\right.\)
Vậy n=0 hoặc n=4
