Question

Tìm n

a) n+6 ⋮ n+1 

b)4n+9⋮2n+1

c)2n⋮n-1

d)n+4⋮n+1

Answer 1

a) Ta có : n+1⋮ n+1

⇒[(n+6)-(n+1)]⋮n+1

⇒5⋮n+1

⇒n+1ϵ {-1;1;5;-5}

⇒nϵ{0;-2;4;-6}

b) Ta có :2(2n+1)⋮2n+1⇔4n+2⋮2n+1

Mà 4n+9⋮2n+1

⇒[(4n+9)-(4n+2)]⋮2n+1

⇒7⋮2n+1⇔2n+1ϵ{-1;1;-7;7}

2n+1	1	-1	-7	7
2n	0	-2	-8	6
n	0	-1	-4	3

c)Ta có : 2(n-1)⋮n-1⇔2n-2⋮n-1

⇒[(2n)-(2n-2)]⋮n-1

⇒2⋮n-1⇔n-1ϵ{1;-1;-2;2}

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

d)n+4⋮n+1

⇒[(n+4)-(n+1)]⋮n+1

⇒3⋮n+1⇔n+1ϵ{1;-1;3;-3}

n+1	1	-1	3	-3
n	0	-2	2	-4