a: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)

b: \(\Leftrightarrow n-1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{2;0;8;-6\right\}\)

a, \(n^2+5=n^2+n-n-1+6=n\left(n+1\right)-\left(n+1\right)+6\)

\(\Rightarrow n+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

b, tương tự 

a: \(3< n^2< 30\)

=>\(\sqrt{3}< n< \sqrt{30}\)

mà \(n\in Z^+\)

nên \(n\in\left\{2;3;4;5\right\}\)

=>A={2;3;4;5}

b: |n|<3

=>-3<n<3

mà \(n\in Z\)

nên \(n\in\left\{-2;-1;0;1;2\right\}\)

=>B={-2;-1;0;1;2}

c: x=3k

=>\(x⋮3\)

mà -4<x<12

nên \(x\in\left\{-3;0;3;6;9\right\}\)

=>C={-3;0;3;6;9}

d: \(n\in N\)

mà n<5

nên \(n\in\left\{0;1;2;3;4\right\}\)

=>\(n^2+3\in\left\{3;4;7;12;19\right\}\)

=>D={3;4;7;12;19}

Ta có với mọi số nguyên m thì m² chia cho 5 dư 0 , 1 hoặc 4.

+ Nếu n² chia cho 5 dư 1 thì   n 2 = 5 k + 1 = > n 2 + 4 = 5 k + 5 ⋮ 5 ; k ∈ N * .

Nên n²+4 không là số nguyên tố

+ Nếu n² chia cho 5 dư 4 thì  n 2 = 5 k + 4 = > n 2 + 16 = 5 k + 20 ⋮ 5 ; k ∈ N * .

Nên n²+16 không là số nguyên tố.

Vậy n² ⋮  5 hay n  ⋮ 5

Đặt \(A=n\left(n^2+1\right)\left(n^2+4\right)\)

TH1: n=5k

\(A=n\left(n^2+1\right)\left(n^2+4\right)\)

\(=5k\left\lbrack\left(5k\right)^2+1\right\rbrack\left\lbrack\left(5k\right)^2+4\right\rbrack\) ⋮5(1)

TH2: n=5k+1

\(A=n\left(n^2+1\right)\left(n^2+4\right)\)

\(=\left(5k+1\right)\left\lbrack\left(5k+1\right)^2+1\right\rbrack\left\lbrack\left(5k+1\right)^2+4\right\rbrack\)

\(=\left(5k+1\right)\left(25k^2+10k+1+1\right)\left(25k^2+10k+1+4\right)\)

\(=\left(5k+1\right)\left(25k^2+10k+2\right)\left(25k^2+10k+5\right)\)

\(=5\left(5k+1\right)\left(25k^2+10k+2\right)\left(5k^2+2k+1\right)\) ⋮5(2)

TH3: n=5k+2

\(A=n\left(n^2+1\right)\left(n^2+4\right)\)

\(=\left(5k+2\right)\left\lbrack\left(5k+2\right)^2+1\right\rbrack\left\lbrack\left(5k+2\right)^2+4\right\rbrack\)

\(=\left(5k+2\right)\left(25k^2+20k+4+1\right)\left(25k^2+20k+4+4\right)\)

\(=\left(5k+2\right)\left(25k^2+20k+5\right)\left(25k^2+20k+8\right)\)

\(=5\left(5k+2\right)\left(5k^2+4k+1\right)\left(25k^2+20k+8\right)\) ⋮5(3)

TH4: n=5k+3

\(A=n\left(n^2+1\right)\left(n^2+4\right)\)

\(=\left(5k+3\right)\left\lbrack\left(5k+3\right)^2+1\right\rbrack\left\lbrack\left(5k+3\right)^2+4\right\rbrack\)

\(=\left(5k+3\right)\left(25k^2+30k+9+1\right)\left(25k^2+30k+9+4\right)\)

\(=\left(5k+3\right)\left(25k^2+30k+10\right)\left(25k^2+30k+13\right)\)

\(=5\left(5k+3\right)\left(5k^2+60k+2\right)\left(25k^2+30k+13\right)\) ⋮5(4)

TH5: n=5k+4

\(A=n\left(n^2+1\right)\left(n^2+4\right)\)

\(=\left(5k+4\right)\left\lbrack\left(5k+4\right)^2+1\right\rbrack\left\lbrack\left(5k+4\right)^2+4\right\rbrack\)

\(=\left(5k+4\right)\left(25k^2+40k+16+1\right)\left(25k^2+40k+16+4\right)\)

\(=\left(5k+4\right)\left(25k^2+40k+17\right)\left(25k^2+40k+20\right)\)

\(=5\left(5k+4\right)\left(25k^2+40k+17\right)\left(5k^2+8k+4\right)\) ⋮5(5)

Từ (1),(2),(3),(4),(5) suy ra A⋮5

c.ơn ck iuu=)) 

\(A=n^2+n+1=n\left(n+1\right)+1\)

Với \(n\inℤ\)thì \(n\left(n+1\right)\)là tích của hai số nguyên liên tiếp nên chia hết cho \(2\).

Do đó \(n\left(n+1\right)\)là số chẵn nên \(A=n\left(n+1\right)+1\)là số lẻ. 

Do đó \(A\)không chia hết cho \(4\).