1: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;4;2;-2;-1;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;3;-3\right\}\)

hay \(n\in\left\{0;1;-1\right\}\)

1: Vì 7 là số nguyên tố nên \(n^7-n⋮7\)

2: \(A=n^3+11n\)

\(=n^3-n+12n\)

\(=n\left(n-1\right)\left(n+1\right)+12n⋮6\)

3: \(=n\left(n^2+3n+2\right)=n\left(n+1\right)\left(n+2\right)⋮6\)

a, Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)

\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)

\(=5n^2+5n=5\left(n^2+n\right)⋮5\)

\(\Rightarrowđpcm\)

b, \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+31n+5-6n^2-7n+5\)

\(=24n+10=2\left(12n+5\right)⋮2\)

\(\Rightarrowđpcm\)

a) $\left(n^2+3n-1\right)\left(n+2\right)-n^3+2$

= n³ + 2n² + 3n² + 6n - n - 2 + 2

= 5n² + 5n

= 5(n² + n ) chia hết cho 5

b) $\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)$

$=\; 6n2\; +\; 30n\; +\; n\; +\; 5\; -\; 6n2\; +\; 3n\; -\; 10n\; +5$

$=\; 24n\; +\; 10$

= 2(12n +5) chia hết cho 2

a: \(\left(n+3\right)^2-\left(n-1\right)^2\)

\(=\left(n+3+n-1\right)\left(n+3-n+1\right)\)

\(=4n\left(2n+2\right)⋮8\)

a,(2n+4).2=4(n+2) chia hwtc ho 8

a) \(\left(n+3\right)^2-\left(n-1\right)^2\)

\(=\left(n+3+n-1\right)\left(n+3-n+1\right)\)

\(=\left(2n+2\right)4\)

\(=2\left(n+1\right).4\)

\(=8\left(n+1\right)⋮8\) 

=> đpcm

a/\(\left(n+3\right)^2-\left(n-1\right)^2.\)

\(=\left(n^2+6n+9\right)-\left(n^2-2n+1\right)\)

\(=n^2+6n+9-n^2+2n-1\)

\(=8n+8\)

\(=8\left(n+1\right)\)

có \(8\left(n+1\right)⋮8\)

\(\Rightarrow\left(n+3\right)^2-\left(n-1\right)^2⋮8\)

b/ \(\left(n+6\right)^2-\left(n-6\right)^2\)

\(=\left(n^2+12n+36\right)-\left(n^2-12n+36\right)\)

\(=n^2+12n+36-n^2+12n-36\)

\(=24n\)

có \(24n⋮24\)

\(\Rightarrow\left(n+6\right)^2-\left(n-6\right)^2⋮24\)

a)

\(55^{n+1}-55^n\\ =55^n.55-55^n\\ =55^n\left(55-1\right)\\ =55^n.54⋮54\\ \RightarrowĐpcm\)

b)

\(n^2\left(n+1\right)+2n\left(n+1\right)\\ =\left(n+1\right)\left(n^2+2n\right)\\ =n\left(n+1\right)\left(n+2\right)⋮6\\ \)

c)

\(2^{n+2}+2^{n+1}+2^n\\ =2^n.2^2+2^n.2+2^n\\ =2^n\left(4+2+1\right)\\ =2^n.7⋮7\)

a) \(\left(5n+7\right)\left(4n+6\right)\)

\(=\left(5n+7\right)4n+\left(5n+7\right)6\)

\(=20n^2+28n+30n+32\)

\(=20n^2+58n+32\)

Vì \(20n^2⋮2\) ; \(58n⋮2\) ; \(32⋮2\) nên \(\left(5n+7\right)\left(4n+6\right)⋮2\)

b) \(\left(8n+1\right)\left(6n+5\right)\)

\(=\left(8n+1\right)6n+\left(8n+1\right)5\)

\(=48n^2+6n+40n+5\)

\(=48n^2+46n+5\)

Vì \(\left(48n^2+46n\right)⋮2\) mà \(5⋮̸2\) nên \(\left(8n+1\right)\left(6n+5\right)⋮̸2\)

c) \(n\left(n+1\right)\left(2n+1\right)\)

\(=n\left(n+1\right)\left(n-1+n-2\right)\)

\(=n\left(n-1\right)\left(n+1\right)+n\left(n+1\right)\left(n+2\right)\)

Với \(\forall n\in N\), tích 3 số tự nhiên liên tiếp chia hết cho 6 nên \(n\left(n-1\right)\left(n+1\right)⋮6\) và \(n\left(n+1\right)\left(n+2\right)⋮6\)

Vậy \(n\left(n+1\right)\left(2n+1\right)⋮6\)