Question

a)  3n+5 chia hết cho n-1

b)  8-3n chia hết cho n+3 

Tìm n:

 

Answer 1

a) Ta có: \(3n+5⋮n-1\)

\(\Rightarrow3.\left(n-1\right)+8⋮n-1\)

\(\Rightarrow8⋮n-1\)

\(\Rightarrow n-1\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow n\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)

Vậy:............

b) \(8-3n⋮n+3\)

\(\Rightarrow3n-8⋮n+3\)

\(\Rightarrow\hept{\begin{cases}3n-8⋮n+3\\n+3⋮n+3\end{cases}}\Rightarrow\hept{\begin{cases}3n-8⋮n+3\\3n+9⋮n+3\end{cases}}\)

\(\Rightarrow\left(3n+9\right)-\left(3n-8\right)⋮n+3\)

\(\Rightarrow17⋮n+3\)

\(\Rightarrow n+3\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)

\(\Rightarrow n\in\left\{-20;-4;-2;14\right\}\)

Vậy:......................