b: Sửa đề: \(n^4-10n^2+9\)
Đặt \(A=n^4-10n^2+9\)
n là số lẻ nên n=2k+1
\(n^4-10n^2+9\)
\(=n^4-n^2-9n^2+9\)
\(=n^2\left(n^2-1\right)-9\left(n^2-1\right)\)
\(=\left(n^2-1\right)\left(n^2-9\right)\)
=(n-1)(n+1)(n-3)(n+3)
=(2k+1-1)(2k+1+1)(2k+1-3)(2k+1+3)
\(=2k\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)=16\cdot k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)
Vì k;k+1;k-1;k+2 là bốn số nguyên liên tiếp
nên k(k+1)(k-1)(k+2)⋮4!=24
=>A⋮16*24
=>A⋮384
c: Đặt \(B=n^6+n^4-2n^2\)
\(=n^2\left(n^4+n^2-2\right)\)
\(=n^2\left(n^4+2n^2-n^2-2\right)\)
\(=n^2\left(n^2+2\right)\left(n^2-1\right)\)
\(=n^2\left(n-1\right)\left(n+1\right)\left(n^2+2\right)\)
\(=n^2\left(n-1\right)\left(n+1\right)\left(n^2-4+6\right)\)
\(=n^2\left(n-1\right)\left(n+1\right)\left(n^2-4\right)+6n^2\left(n-1\right)\left(n+1\right)\)
\(=n^2\left(n-1\right)\left(n+1\right)\left(n-2\right)\left(n+2\right)+6n^2\left(n-1\right)\left(n+1\right)\)
\(=n\left(n-1\right)\left(n-2\right)\cdot n\left(n+1\right)\left(n+2\right)+6n^2\left(n-1\right)\left(n+1\right)\)
*Chứng minh \(n^2\left(n^2-1\right)\left(n^2-4\right)\) ⋮72
Vì n;n-1;n-2 là 3 số nguyên liên tiếp
nên n(n-1)(n-2)⋮3!=6(2)
Vì n;n+1;n+2 là ba số nguyên liên tiếp
nên n(n+1)(n+2)⋮3!=6(1)
Từ (1),(2) suy ra n(n-1)(n-2)*n(n+1)(n+2)⋮6*6
=>\(n^2\left(n^2-1\right)\left(n^2-4\right)\) ⋮36
Vì n;n-1;n-2;n+1 là bốn số nguyên liên tiếp
nên n(n-1)(n-2)(n+1)⋮4!=24
=>\(n^2\left(n^2-1\right)\left(n^2-4\right)\) ⋮24
mà \(n^2\left(n^2-1\right)\left(n^2-4\right)\) ⋮36
và BCNN(24;36)=72
nên \(n^2\left(n^2-1\right)\left(n^2-1\right)\) ⋮72(3)
*Chứng minh \(6n^2\left(n-1\right)\left(n+1\right)\) ⋮72
TH1: n=2k
\(6n^2\left(n-1\right)\left(n+1\right)=6\cdot\left(2k\right)^2\left(2k-1\right)\left(2k+1\right)\)
\(=6\cdot2k\cdot2k\left(2k-1\right)\left(2k+1\right)=12k\cdot2k\left(2k-1\right)\cdot\left(2k+1\right)\)
Vì 2k;2k-1;2k+1 là ba số nguyên liên tiếp
nên 2k(2k-1)(2k+1)⋮3!=6
=>\(12k\cdot2k\left(2k-1\right)\left(2k+1\right)\) ⋮6*12
=>\(6n^2\left(n-1\right)\left(n+1\right)\) ⋮72(4)
TH2: n=2k+1
\(6n^2\left(n-1\right)\left(n+1\right)\)
\(=6\left(2k+1\right)^2\cdot\left(2k+1-1\right)\left(2k+1+1\right)\)
\(=6\left(2k+1\right)^2\cdot2k\cdot\left(2k+2\right)\)
\(=24k\left(k+1\right)\left(2k+1\right)^2\) ⋮24
\(k\left(k+1\right)\left(2k+1\right)=k\left(k+1\right)\left(k+2+k-1\right)\)
=k(k+1)(k+2)+k(k+1)(k-1)
Vì k;k+1;k+2 là ba số nguyên liên tiếp
nên k(k+1)(k+2)⋮3!=6(5)
Vì k;k+1;k-1 là ba số nguyên liên tiếp
nên k(k+1)(k-1)⋮3!=6(6)
Từ (5),(6) suy ra k(k+1)(k+2)+k(k+1)(k-1)⋮6
=>k(k+1)(2k+1)⋮6
=>\(24k\left(k+1\right)\left(2k+1\right)\cdot\left(2k+1\right)\) ⋮6*24=144
=>2k(k+1)(2k+1)^2⋮72
=>\(6n^2\left(n-1\right)\left(n+1\right)\) ⋮72(7)
Từ (4),(7) suy ra \(6n^2\left(n-1\right)\left(n+1\right)\) ⋮72(8)
Từ (3),(8) suy ra \(=n\left(n-1\right)\left(n-2\right)\cdot n\left(n+1\right)\left(n+2\right)+6n^2\left(n-1\right)\left(n+1\right)\) ⋮72
=>\(n^6+n^4-2n^2\) ⋮72
