[Toán 8] Toán nâng cao đại số 8

V

vipboycodon

$x^{200}+x^{100}+1$

$= x^{200}-x^8+x^{100}-x^4+x^8+x^4+1$

$= x^8(x^{192}-1)+x^4(x^{96}-1)+x^8+x^4+1$

$= x^8[(x^6)^{32}-1]+x^4[(x^6)^{16}-1]+x^8+x^4+1$

Vì $(x^6)^{32}-1 \ \vdots \ x^6-1$ , $(x^6)^{16}-1 \ \vdots \ x^6-1$ , $x^8+x^4+1 = (x^4-x^2+1)(x^4+x^2+1) \ \vdots \ x^4+x^2+1$ nên ta có đpcm.
 
You must log in or register to reply here.
Top Bottom