Vì : \(2n+7⋮n+1\) (1)
Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\) (2)
Từ (1) và (2) \(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{\pm1;\pm5\right\}\Rightarrow n\in\left\{0;-2;4;-6\right\}\)
Vậy \(n\in\left\{0;-2;4;-6\right\}\)
2n + 7 = 2n + 2 + 5 = 2(n + 1) + 5 \(⋮\) n + 1
=> 5 \(⋮\) n + 1 => n + 1 \(\in\) Ư(5)
=> n + 1 \(\in\) {-5; -1; 1; 5}
=> n \(\in\) {-6; -2; 0; 4}
(ĐK : n \(\in\) N)
Ta có:
(2n + 7) \(⋮\) (n + 1)
\(\Rightarrow\) (2n + 2 + 5) \(⋮\) (n + 1)
\(\Leftrightarrow\) [2.(n + 1) + 5] \(⋮\) (n + 1)
Vì (n + 1) \(⋮\) (n + 1) \(\Rightarrow\) 2.(n + 1) \(⋮\) (n + 1)
\(\Rightarrow\) 5 \(⋮\) (n + 1)
\(\Rightarrow\) n + 1 \(\in\) Ư (5)
\(\Rightarrow\) n + 1 \(\in\) {1 ; 5}
n + 1 = 1
n = 1 -1
n = 0
n + 1 = 5
n = 5 - 1
n = 4
Vậy n \(\in\) {0 ; 4}
Nhớ tick ủng hộ mình nhé!
\(\left(2n+7\right)⋮\left(n+1\right)\)
\(\Rightarrow2n+2+5⋮n+1\)
\(\Rightarrow2\left(n+1\right)+5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
nếu n+1=1 thì n=0
nếu n+1=-1 thì n=-2
nếu n+1=5 thì n=4
nếu n+1=-5 thì n=-6
Vậy \(x\in\left\{0;-2;4;-6\right\}\)
Ta có 2n+7 ⋮ n+1(1)
n+1⋮n+1=>2(n+1)⋮n+1=>2n+2⋮n+1(2)
Từ (1) và (2)=> (2n+7)-(2n+2)⋮n+1 => 5⋮n+1 => n+1 thuộc Ư(5)={1;-1;5;-5}
=>n thuộc {0;-2;4;-6}
Vậy n thuộc {0;-2;4;-6}
\(\left(2n+7\right)⋮\left(n+1\right)\)
\(Có\)\(\left(n+1\right)⋮\left(n+1\right)\)
\(\Rightarrow2\left(n+1\right)⋮\left(n+1\right)\)
\(\Rightarrow2n+2⋮n+1\)
Mà \(\left(2n+7\right)⋮\left(n+1\right)\)
\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)\)
\(\Rightarrow2n+7-2n-2\)
\(\Rightarrow5⋮\left(n+1\right)\)
\(\Rightarrow n+1\inƯ\left(5\right)\)
Ư(5)=\(\left\{\pm1;\pm5\right\}\)
n+1 \(\in\)\(\left\{\pm1;\pm5\right\}\)
n\(\in\)\(\left\{0;-2;4;-6\right\}\)
Vậy n=0 n=4
