Toán x mũ 16 – 1 giúp mình vs ạ Phân Tích Đa Thức Thành Nhân Tử 05/07/2021 By Arianna x mũ 16 – 1 giúp mình vs ạ Phân Tích Đa Thức Thành Nhân Tử
#Ban tham khao: x^16 – 1 ( x^8)² – 1² = ( x^8 – 1) ( x^8 + 1) =[ (x^4)² – 1² ] ( x^8 + 1) = ( x^4 – 1) (x^4 + 1) ( x^8 + 1) = [ ( x² )² – 1² ] ( x^4 + 1) ( x^8 + 1) =( x² – 1²) ( x² + 1) ( x^4 + 1) ( x^8 + 1) = ( x- 1) ( x+ 1) ( x² + 1 ) ( x^4 + 1) (x^8 + 1) Chuc ban hoc tot. Trả lời
$x^{16}-1$ $⇔(x^8)^2-1^2_{}$ $⇔(x^8-1)(x^8+1)_{}$ $⇔[ (x^4)^2-1^2].(x^8+1)_{}$ $⇔(x^4-1)(x^4+1)(x^8+1)_{}$ $⇔[ (x^2)^2-1^2].(x^4+1)(x^8+1)_{}$ $⇔(x^2-1^2)(x^2+1)(x^4+1)(x^8+1)_{}$ $⇔(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)_{}$ Trả lời
#Ban tham khao:
x^16 – 1 ( x^8)² – 1² = ( x^8 – 1) ( x^8 + 1)
=[ (x^4)² – 1² ] ( x^8 + 1)
= ( x^4 – 1) (x^4 + 1) ( x^8 + 1)
= [ ( x² )² – 1² ] ( x^4 + 1) ( x^8 + 1)
=( x² – 1²) ( x² + 1) ( x^4 + 1) ( x^8 + 1)
= ( x- 1) ( x+ 1) ( x² + 1 ) ( x^4 + 1) (x^8 + 1)
Chuc ban hoc tot.
$x^{16}-1$
$⇔(x^8)^2-1^2_{}$
$⇔(x^8-1)(x^8+1)_{}$
$⇔[ (x^4)^2-1^2].(x^8+1)_{}$
$⇔(x^4-1)(x^4+1)(x^8+1)_{}$
$⇔[ (x^2)^2-1^2].(x^4+1)(x^8+1)_{}$
$⇔(x^2-1^2)(x^2+1)(x^4+1)(x^8+1)_{}$
$⇔(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)_{}$