toán chuyên

T

transformers123

$x^8-1$

$=(x^4-1)(x^4+1)$

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

Dễ thấy $(x^2-1)(x^2+1)(x^4+1)\ \vdots\ x^2+1$

$\Longrightarrow x^8-1\ \vdots\ x^2+1$
 
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