\(2n-3⋮n+1\)
\(\Rightarrow\left(2n+2\right)-2-3⋮n+1\)
\(\Rightarrow2\left(n+1\right)-5⋮n+1\)
\(2\left(n+1\right)⋮n+1\)
\(\Rightarrow-5⋮n+1\)
\(\Rightarrow\) \(n+1\inƯ\left(-5\right)\)
đến đây dễ r`, bn tự lm tiếp đi!
\(2n-3⋮n+1\)
\(\Rightarrow2n+2-5⋮n+1\)
mà \(2n+2⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)\)
\(\Rightarrow n+1\in\left\{1;\left(-1\right);5;\left(-5\right)\right\}\)
\(\Rightarrow n\in\left\{0;\left(-2\right);\left(-6\right);4\right\}\)
vậy :n = 0
n = -2
n = -6
n = 4
2n - 3 \(⋮\)n + 1
=> 2n + 2 - 5 \(⋮\)n + 1
=> 2 ( n + 1 ) - 5 \(⋮\)n + 1
Ta thấy 2 ( n + 1 ) \(⋮\)n + 1
=> 5 \(⋮\)n + 1
=> n + 1 \(\in\)Ư ( 5 )
Ư ( 5 ) = { 1 ; - 1 ; 5 ; - 5 )
Ta có bảng sau :
| n + 1 | 1 | - 1 | 5 | - 5 |
| n | 0 | - 2 | 4 | - 6 |
Vậy .....
\(2n-3⋮n+1\)
ta có \(n+1⋮n+1\)
\(\Rightarrow2\left(n+1\right)⋮n+1\)
\(\Rightarrow2n+2⋮n+1\)
mà \(2n-3⋮n+1\)
\(\Rightarrow2n+2-\left(2n-3\right)⋮n+1\)
\(\Rightarrow2n+2-2n+3\) \(⋮n+1\)
\(\Rightarrow5\) \(⋮n+1\)
\(\Rightarrow n+1\in\text{Ư}_{\left(5\right)}\)
\(\text{Ư}_{\left(5\right)}=\text{ }\left\{1;-1;5;-5\right\}\)
lập bảng giá trị
| \(n+1\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(n\) | \(0\) | \(-2\) | \(4\) | \(-6\) |
vậy..................
