Question

cho x thuộc Z cmr :x²⁰⁰+x¹⁰⁰+1 chia hết cho x⁴+x²+1

Answer 1

Lời giải:

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

$=x^2(x^{198}-1)+x^4(x^{96}-1)+x^4+x^2+1$

Ta thấy:

$x^{198}-1=(x^6)^{33}-1^{33}\vdots x^6-1=(x^2-1)(x^4+x^2+1)\vdots x^4+x^2+1$

$x^{96}-1=(x^6)^{16}-1\vdots x^6-1=(x^2-1)(x^4+x^2+1)\vdots x^4+x^2+1$

$x^4+x^2+1\vdots x^4+x^2+1$

$\Rightarrow x^2(x^{198}-1)+x^4(x^{96}-1)+x^4+x^2+1\vdots x^4+x^2+1$ (đpcm)