Question

tìm số nguyên n thỏa mãn 2n + 1 chia hết cho n-2

Answer 1

Ta có \(\hept{\begin{cases}2n+1⋮n-2\\n-2⋮n-2\end{cases}\Rightarrow\hept{\begin{cases}2n+1⋮n-2\\2n-4⋮n-2\end{cases}}}\)

\(\Rightarrow2n+1-2n+4⋮n-2\)

\(\Rightarrow5⋮n-2\)

\(\Rightarrow n-2\in\left\{1;5\right\}\)

\(\Rightarrow n\in\left\{3;7\right\}\)

Answer 2

Ta có:  2n+1\(⋮\)n-2

\(\Rightarrow\)2n-4+5\(⋮\)n-2

\(\Rightarrow\)2(n-2)+5\(⋮\)n-2

Mà 2(n-2)\(⋮\)n-2                   (\(\forall\)n\(\in\)Z)

Nên 5\(⋮\)n-2

  n-2\(\in\)Ư(5)=\([\)-1;1;5;-5\(]\)(dấu ngoặc sai nhé)

n\(\in\)\([\)1;3;7;-3\(]\)

Answer 3

\(2n+1⋮n-2\)

\(\Rightarrow2\left(n-2\right)+5⋮n-2\)

\(\Rightarrow5⋮n-2\)

\(\Rightarrow n-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow n\in\left\{3;1;7;-3\right\}\)

Vậy.............................