[Toán 8] Thu gọn $\dfrac{x^{10}-x^8-x^7+x^6+x^5+x^4-x^3-x^2+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^6+1} $

M

mikelhpdatke

Ta có:
$A=x^{10}-x^8-x^7+x^6+x^5+x^4-x^3-x^2+1$

$=(x^4-x^2+1)(x^6-x^3+1)$

$B=x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$

$=(x^2+1)(x^4-x^2+1)(x^6-x^3+1)((x^6+x^3+1)(x^{12}-x^6+1)$

$\rightarrow \dfrac{A}{B} = \dfrac{(x^4-x^2+1)(x^6-x^3+1)}{(x^2+1)(x^4-x^2+1)(x^6-x^3+1)(x^6+x^3+1)(x^{12}-x^6+1)}$

$=\dfrac{1}{(x^2+1)(x^6+x^3+1)(x^{12}-x^6+1)}$

PS: Lỗi tatex sao mà mình sửa mãi ko được, kq là 1/.....
 
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