Câu hỏi của Cỏ dại - Toán lớp 8 - Học toán với OnlineMath

a)Đặt \(A=n^3+6n^2+8n\)

\(A=n\left(n^2+6n+8\right)\)

\(A=n\left(n^2+2n+4n+8\right)\)

\(A=n\left[n\left(n+2\right)+4\left(n+2\right)\right]\)

\(A=n\left(n+2\right)\left(n+4\right)⋮\forall n\) chẵn

b)Đặt \(B=n^4-10n^2+9\)

\(B=n^4-n^2-9n^2+9\)

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

\(B=\left(n-3\right)\left(n-1\right)\left(n+1\right)\left(n+3\right)⋮384\forall n\) lẻ

\(b,n^4-10n^2+9=n^4-n^2-9n^2+9=\left(n^2-1\right)\left(n^2-9\right)\\ =\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)\)

Vì \(n\in Z\) và n lẻ nên \(n=2k+1\left(k\in Z\right)\)

\(\Leftrightarrow\left(n-1\right)\left(n+1\right)\left(n-3\right)\left(n+3\right)\\ =2k.\left(2k+2\right).\left(2k-2\right).\left(2k+4\right)\\ =16k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)

Vì \(k,k+1,k-1,k+2\) là 4 số nguyên liên tiếp nên chia hết cho \(1.2.3.4=24\)

Do đó \(16k\left(k+1\right)\left(k-1\right)\left(k+2\right)⋮24.16=384\)

\(c,\forall n=1\Leftrightarrow10+18-28=0⋮27\\ \text{G/s }n=k\Leftrightarrow\left(10^k+18k-28\right)⋮27\\ \Leftrightarrow10^k+18k-28=27m\left(m\in N\right)\\ \Leftrightarrow10^k=27m-18k+28\\ \forall n=k+1\Leftrightarrow10^{k+1}+18\left(k+1\right)-28\\ =10.10^k+18k-10\\ =10\left(27m-18k+28\right)+18k-10=270m-162k+270⋮27\)

Theo PP quy nạp ta đc đpcm

\(n^4-10n^2+9=\left(n^4-9n^2\right)-\left(n^2-9\right)\)

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

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

Vì n lẻ \(\Rightarrow n=2k+1\)( \(k\inℤ\))

\(\Rightarrow n^4-10n^2+9=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+3\right)\)

\(=2k.\left(2k+2\right).\left(2k-2\right).\left(2k+4\right)\)

\(=16.k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)

\(=16.\left(k-1\right).k.\left(k+1\right).\left(k+2\right)\)

Vì \(k-1\); \(k\); \(k+1\); \(k+2\)là 4 số nguyên liên tiếp

\(\Rightarrow\left(k-1\right).k.\left(k+1\right).\left(k+2\right)⋮24\)

\(\Rightarrow16.\left(k-1\right).k.\left(k+1\right).\left(k+2\right)⋮384\)

hay \(n^4-10n^2+9⋮384\)( đpcm )

A=(n^2-9)(n^2-1)

=(n-3)(n+3)(n-1)(n+1)

=(2k+1-3)(2k+1+3)(2k+1-1)(2k+1+1)

=2k(2k+2)(2k-2)(2k+4)

=16k(k+1)(k-1)(k+2)

Vì k;k+1;k-1;k+2là 4 số liên tiếp

nen k(k-1)(k+1)(k+2) chia hết cho 4!=24

=>A chia hết cho 384

\(n^4-10n^2+9\)

\(=\)\(\left(n^4-n^2\right)-\left(9n^2-9\right)\)

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

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

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

Mà n lẻ nên n có dạng \(2k+1\) \(\left(k\inℤ\right)\)

\(=\)\(\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1-3\right)\left(2k+1+3\right)\)

\(=\)\(2k\left(2k+2\right)\left(2k-2\right)\left(2k+4\right)\)

\(=\)\(16k\left(k+1\right)\left(k-1\right)\left(k+2\right)\)

\(=\)\(15k\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Lại có : 

\(16k\left(k+1\right)\left(k-2\right)\left(k+2\right)⋮16\)

\(15\left(k-1\right)k\left(k+1\right)\left(k+2\right)⋮8,⋮3\)

\(\Rightarrow\)\(15\left(k-1\right)k\left(k+1\right)\left(k+2\right)⋮384\) ( đpcm ) 

Vậy \(n^4-10n^2+9⋮384\) với mọi n là số nguyên lẻ 

Chúc bạn học tốt ~