Question

a) n+3 chia hết cho n-1

b)4n+3 chia hết cho 2n+1

Answer 1

a) n+3 ⋮ n-1

=> (n+3)-(n-1) ⋮(n-1)

=> (n+3-n+1) ⋮(n-1)

=> 4 ⋮(n-1)

=> (n-1) ∈ Ư(4)= { -4;-2;-1;1;2;4}

=> n ∈ { -3,-1;0;2;3;5}

b)4n+3 ⋮ 2n+1

vì (2n+1) ⋮ 2n+1

=> 2(2n+1) ⋮ 2n+1

=> (4n+2) ⋮ 2n+1

=> (4n+3)-(4n+2) ⋮ 2n+1

=> (4n+3-4n-2) ⋮ 2n+1

=> 1 ⋮ 2n+1

=> (2n+1) ∈ Ư(1)={-1;1}

ta có bảng sau

2n+1	-1	1
n	-1	0

vậy n ∈ {-1;0}