Question

Tìm n ∈ Z sao cho 2n – 3 chia hết cho n + 1.

Answer 1

Ta có: 2n-3 = (2n+2)-2-3 = 2(n+1)-5

Để 2n-3 \(⋮\) n+1 thì -5 \(⋮\) n+1

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

Ta có bảng sau:

n+1	-5	-1	1	5
n	-6	-2	0	4

Vậy n \(\in\left\{-6;-2;0;4\right\}\)

Answer 2

Ta có:

\(2n-3⋮n+1\)

=>\(\text{2.(n+1)-5}⋮n+1\)

=>\(5⋮n+1\) ( \(2.\left(n+1\right)⋮n+1\) )

=>\(n+1\inƯ\left(5\right)\) \(\left(n\in Z\right)\)

Ta có bảng sau:

n+1	1	-1	5	-5
n	0	-2	4	-6

Vậy \(n\in\left\{0;-2;4;-6\right\}\)