Question

Cho A = 5n +1/n+1 ( n khác -1 ) . Tìm n thuộc N  để A nguyên 

Answer 1

\(A=\frac{5n+1}{n+1}=\frac{5n+5-4}{n+1}=\frac{5\left(n+1\right)-4}{n+1}=5+\frac{-4}{n+1}\)

Để \(A\inℤ\Leftrightarrow-4⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4​\right\}\)

\(\Leftrightarrow n\in\left\{0;-2;1;-3;3;-5​\right\}\)

Answer 2

nếu để A nguyên thì ta có 5n+1( ba chấm dọc)n+1

=)5(n+1)-4 : n+1

=) 4: n+1

=) n+1 thuộc Ự(4)=(1;2;4)