Question

[Ngô Châu Bảo Oanh]

CMR:

a, x⁸ⁿ+x⁴ⁿ+1\(⋮\)x²ⁿ+xⁿ+1

b,x^3m+1+x^3n+2\(⋮\)x²+x+1\(\forall\)m, n \(\in\)N

Answer 1

[Akai Haruma]

Lời giải:

Ta có:

\(x^{8n}+x^{4n}+1=(x^{4n})^2+2.x^{4n}+1-x^{4n}\)

\(=(x^{4n}+1)^2-x^{4n}=(x^{4n}+1+x^{2n})(x^{4n}+1-x^{2n})\)

Xét \(x^{4n}+1+x^{2n}=(x^{2n})^2+2.x^{2n}+1-x^{2n}=(x^{2n}+1)^2-x^{2n}\)

\(=(x^{2n}+1+x^n)(x^{2n}+1-x^n)\)

Do đó:

\(x^{8n}+x^{4n}+1=(x^{4n}+1-x^{2n})(x^{2n}+1+x^n)(x^{2n}+1-x^n)\)

\(\Rightarrow x^{8n}+x^{4n}+1\vdots x^{2n}+x^n+1\) (đpcm)

b)

Sửa đề: \(x^{3m+1}+x^{3n+2}+1\vdots x^2+x+1\)

Đặt \(A=x^{3m+1}+x^{3n+2}+1\)

\(\Leftrightarrow A=x(x^{3m}-1)+x+x^2(x^{3n}-1)+x^2+1\)

\(\Leftrightarrow A=x[ (x^3)^m-1]+x^2[(x^3)^n-1]+(x^2+x+1)\)

Khai triển:

\((x^3)^m-1=(x^3)^m-1^m=(x^3-1).T=(x-1)(x^2+x+1)T\)

(đặt là T vì phần biểu thức đó không quan trọng)

\(\Rightarrow (x^3)^m-1\vdots x^2+x+1\)

Tương tự, \((x^3)^n-1\vdots x^2+x+1\)

Do đó, \(A=x(x^{3m}-1)+x^2(x^{3n}-1)+x^2+x+1\vdots x^2+x+1\)

Ta có đpcm.