CÂU NÀY MÌNH LÀM ĐƯỢC RỒII

Ta có: \(\left(x^{200}+x^{100}+1\right)=\left(x^{100}+1\right)^2\)

\(\left(x^4+x^2+1\right)=\left(x^2+1\right)^2\)

Vì \(1⋮1;x^{100}⋮x^2\forall x\)

\(\Rightarrow x^{100}+1⋮x^2+1\forall x\)

\(\Rightarrow Vớix\in Z,\left(x^{200}+x^{100}+1\right)⋮\left(x^4+x^2+1\right)\)

 \(A=x^{200}+x^{100}+1\)

    \(=x^{200}-x^2+x^{100}-x^4+x^4+x^2+1\)

    \(=x^2\left(x^{198}-1\right)+x^4\left(x^{96}-1\right)+\left(x^4+x^2+1\right)\)

    \(=x^2\left(x^{^6}-1\right).A+x^4\left(x^6-1\right).B+x^4+x^2+1\)

\(x^6-1=\left(x^3-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)=\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)\)

Vậy \(A⋮\left(x^4+x^2+1\right)\)

x^200+x^100+1=x^100*(x^2+1)+1
x^4+x^2+1=x^2*(x^2+1)+1
mà x^100chia hết cho x^2
x^2+1chia hết cho x^2+1
1 chia hết cho1
suy ra x^100*(x^2+1)+1 chia hết cho x^2*(x^2+1)+1 hay x^200+x^100+1 chia hết cho x^4+x^2+1

https://vn.answers.yahoo.com/question/index?qid=20111212062832AACt3bZ

Ta có : x⁶ⁿ-1=(x⁶-1).A=(x²-1)(x⁴+x²+1)A chia hết cho x⁴ + x² +1

Khi đó : M=x²⁰⁰+x¹⁰⁰+1=x²⁰⁰-x²+x¹⁰⁰-x⁴+(x⁴+x²+1)= x²​​[(x⁶)³³-1]-x⁴ [(x⁶)¹⁶-1]+(x⁴ + x² +1)

Vì x²​​[(x⁶)³³-1]chia hết cho x⁴ + x² +1

x⁴ [(x⁶)¹⁶-1]chia hết cho x⁴ + x² +1

Nên .....

làm thế thì hơi rối