\(n+10⋮n-1\)
\(\Rightarrow\left(n-1\right)+11⋮n-1\)
Mà \(\left(n-1\right)⋮n-1\)
\(\Rightarrow11⋮n-1\)
\(\Rightarrow n-1\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow n-1\in\left\{1;-1;11;-11\right\}\)
\(\Rightarrow n\in\left\{2;0;12;-10\right\}\)
Vậy n = -10 ; 0 ;2 ; 12
a,n +10 là bội của n- 1
⇒n +10 ⋮n- 1
⇒n- 1 +11⋮n- 1
Mà n- 1⋮n- 1 nên 11 ⋮n- 1
⇒n- 1 ∈Ư(11) ={1;-1;-11;11}
⇒n- 1 ∈{1;-1;-11;11}
⇒n ∈{2;0;-10;12}
Vậy n ∈{2;0;-10;12}
Ta có: n + 10 là bội của n - 1
\(\Rightarrow n+10⋮n-1\)
\(\Rightarrow\left(n-1\right)+11⋮n-1\)
Vì \(n-1⋮n-1\)
\(\Rightarrow11⋮n-1\)
\(\Rightarrow n-1\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow n\in\left\{2;0;12;-10\right\}\)
n + 10 là bội của n - 1
<=> n + 10 \(⋮\) n - 1
hay n - 1 + 11 \(⋮\) n - 1
vì n - 1 \(⋮\) n - 1 nên để n - 1 + 11 \(⋮\) n - 1 thì
11 \(⋮\)n - 1
=>n - 1 \(\varepsilon\) Ư(11)={-1;1;-11;11}
=> n \(\varepsilon\){-2;0;-12;10}
\(n+10\in B\left(n-1\right)\)
\(\Rightarrow\frac{n+10}{n-1}\)
\(n+10=\left(n-1\right)+11\)
VÌ :
\(n+10⋮n-1\)
\(1\left(n-1\right)⋮n-1\)
\(\Rightarrow11⋮n-1\)
\(\Rightarrow n-1\inƯ\left(11\right)\)
\(\Rightarrow n-1\in\left\{1;-1;11;-11\right\}\)
\(\Rightarrow n\in\left\{2;0;12;-10\right\}\)
VẬY \(n\in\left\{2;0;12;-10\right\}\)
n = 2;0;12;-10