Question

tìm n thuộc Z sao cho n+10 là bội của n-1

Answer 1

\(n+10⋮n-1\)

\(\Rightarrow\left(n-1\right)+11⋮n-1\)

\(\hept{\begin{cases}n-1⋮n-1\\11⋮n-1\end{cases}}\)

\(\Rightarrow n-1\inƯ\left(11\right)\)

\(n-1\in\left\{1;11;-1;-11\right\}\)

\(n\in\left\{2;12;0;-10\right\}\)

Answer 2

\(n+10⋮n-1\)

\(\Rightarrow n-1+11⋮n-1\)

\(\Rightarrow11⋮n-1\)

\(\Rightarrow n-1\inƯ\left(11\right)=\left\{-1;11;-11;1\right\}\)

\(\Rightarrow n\in\left\{0;12;-10;2\right\}\)