a: Vì 5 là số nguyên tố
nên \(n^5-n⋮5\)
b: \(M=2n^4-7n^3-2n^2+13n+6\)
\(=2n^4-4n^3-3n^3+6n^2-8n^2+16n-3n+6\)
\(=2n^3\left(n-2\right)-3n^2\left(n-2\right)-8n\left(n-2\right)-3\left(n-2\right)\)
\(=\left(n-2\right)\left(2n^3-3n^2-8n-3\right)\)
\(=\left(n-2\right)\left(2n^3-6n^2+3n^2-9n+n-3\right)\)
\(=\left(n-2\right)\left[2n^2\left(n-3\right)+3n\left(n-3\right)+\left(n-3\right)\right]\)
\(=\left(n-2\right)\left(n-3\right)\left(2n^2+3n+1\right)\)
\(=\left(n-2\right)\left(n-3\right)\left(n+1\right)\left(2n+1\right)\)
Vì n là số nguyên nên n có thể chia 3 dư 0;1;2
TH1: n chia 3 dư 0
=>n=3k(k\(\in\)Z)
=>n-3=3k-3=3(k-1) chia hết cho 3
=>\(\left(n-2\right)\left(n-3\right)\left(n+1\right)\left(2n+1\right)⋮3\)
=>M\(⋮\)3(1)
TH2: n=3k+1(k\(\in\)Z)
\(2n+1=2\left(3k+1\right)+1=6k+2+1=6k+3=3\left(2k+1\right)⋮3\)
=>\(\left(n-2\right)\left(n-3\right)\left(n+1\right)\left(2n+1\right)⋮3\)
=>\(M⋮3\left(2\right)\)
TH3: n=3k+2(k\(\in\)Z)
\(n+1=3k+2+1=3k+3=3\left(k+1\right)⋮3\)
=>\(\left(n-2\right)\left(n-3\right)\left(n+1\right)\left(2n+1\right)⋮3\)
=>\(M⋮3\left(3\right)\)
Từ (1),(2),(3) suy ra \(M⋮3\)
Vì n-2;n-3 là hai số nguyên liên tiếp
nên \(\left(n-2\right)\left(n-3\right)⋮2\)
=>\(\left(n-2\right)\left(n-3\right)\left(n+1\right)\left(2n+1\right)⋮2\)
=>\(M⋮2\)
mà \(M⋮3\)
và ƯCLN(2;3)=1
nên M chia hết cho 2*3=6
