Question

Bài 1: Tìm số nguyên N ∈ Z sao cho :

a) \(\left(n+5\right):\left(n+2\right)\)

b) \(2.\left(n-1\right)+2:\left(n-1\right)\)

Answer 1

a) \(\left(n+5\right)⋮\left(n+2\right)\)

Ta có : \(n+5=\left(n+2\right)+3\)

Mà \(\left(n+2\right)⋮\left(n+2\right)\)

\(\Rightarrow\) Để \(\left(n+5\right)⋮\left(n+2\right)\) thì 3 phải chia hết cho (n + 2)

\(\Rightarrow\) \(\left(n+2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Ta có bảng sau :

\(n+2\)	\(1\)	\(-1\)	\(3\)	\(-3\)
\(n\)	\(-1\)	\(-3\)	\(1\)	\(-5\)

Vậy \(n\in\left\{-1;-3;1;-5\right\}\)

b) \(2\left(n-1\right)+2⋮\left(n-1\right)\)

Ta có : \(2\left(n-1\right)⋮\left(n-1\right)\)

Để \(2\left(n-1\right)+2⋮\left(n-1\right)\) \(\Rightarrow2⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng sau :

n - 1	1	-1	2	-2
n	2	0	3	-1

Vậy \(n\in\left\{2;0;3;-1\right\}\)