\(n^2+4⋮n+1\)
\(n+1⋮n+1\) nên \(\left(n+1\right)\left(n+1\right)⋮\left(n+1\right)\)
Hay \(n^2+2n+1⋮n+1\)
\(\Rightarrow\left(n^2+2n+1\right)-\left(n^2+4\right)⋮n-1\)
\(\Rightarrow2n+1-4⋮n-1\)
\(\Rightarrow2n-3⋮n-1\)
\(n-1⋮n-1\) nên \(n-2⋮n-1\)
\(\Rightarrow\left(2x-2\right)-\left(2n-3\right)⋮n-1\)
\(\Rightarrow1⋮n-1\)
\(\Rightarrow n-1\in\text{Ư}\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow n\in\left\{0;2\right\}\)
Lấy n2+4 chia n+1 theo cột dọc, ta được
Ta được: n-1 (dư 5)
Ta có \(A=Q+\dfrac{R}{B}=n+1+\dfrac{5}{n+1}\)
\(n+1\inƯ\left(5\right)=\left(\pm1;\pm5\right)\)
Kẻ bảng
| n+1 | 1 | -1 | 5 | -5 |
| n | 0 | -2 | 4 | -6 |
Vì \(n\in N\)* nên \(n=4\left(Nhận\right)\)
Ta có: \(n^2+4\) \(⋮\) \(n+1\) ( n \(\in\) N* )
\(\Leftrightarrow\) n.n + n - n + 4 \(⋮\) n + 1
\(\Leftrightarrow\) n ( n + 1 ) - n + 1 + 3 \(⋮\) n + 1
Vì n ( n + 1 ) \(⋮\) n + 1 và (-n + 1) \(⋮\) n + 1 nên 3 \(⋮\) n + 1
\(\Leftrightarrow\) n + 1 \(\in\) Ư ( 3 ) = { 1 ; 3 }
\(\Leftrightarrow\) n \(\in\) { 0 ; 2 }
Vì n \(\in\) N* nên n ko thể bằng 0
Vậy n = 2
Chúc bạn học tốt!!!
Ta có : \(n^2+4⋮n+1\)
\(\Rightarrow n.n+n-n+4⋮n +1\)
\(\Rightarrow n.\left(n+1\right)+4⋮n+1\)
\(\Rightarrow n.\left(n+1\right)-n+1+3⋮n+1\)
\(\Rightarrow n.\left(n+1\right)-1.\left(n+1\right)+3⋮n+1\)
Vì \(n.\left(n+1\right)⋮n+1\) ; \(1.\left(n+1\right)⋮n+1\)
\(\Rightarrow3⋮n+1\)
\(\Rightarrow n+1\inƯ\left(3\right)\)
\(\Rightarrow n+1\in\left\{-1;1;-3;3\right\}\)
\(\Rightarrow n\in\left\{-2;0;-4;2\right\}\)
Mà \(n\in N^{ }\)*
\(\Rightarrow n=2\)
Vậy n=2
!!! Nhanh thì tickkkkk 