phaan tích đa thức x^10-1 thành nhân tử
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^{10}+x^2+1\)
\(=x^{10}-x^8+x^4+x^8-x^6+x^2+x^6-x^4+1\)
\(=x^4\left(x^6-x^4+1\right)+x^2\left(x^6-x^4+1\right)+\left(x^6-x^4+1\right)\)
\(=\left(x^6-x^4+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^6-x^4+1\right)\left[x^4+2x^2+1-x^2\right]\)
\(=\left(x^6-x^4+1\right)\left[\left(x^2+1\right)^2-x^2\right]\)
\(=\left(x^6-x^4+1\right)\left(x^2+1+x\right)\left(x^2+1-x\right)\)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x^10+x^5+1
=x10-x+x5-x2+x2+x+1
=x.(x9-1)+x2.(x3-1)+(x2+x+1)
=x.(x3-1)(x3+1)+x2(x3-1)+(x2-x+1)
=x.(x-1)(x2+x+1)(x3+1)+x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x.(x-1)(x3+1)+x2(x-1)+1]
=(x2+x+1)(x5+x2-x4-x+x3-x2+1)
=(x2+x+1)(x5-x4+x3-x+1)
x^10+x^5+1
=x10-x+x5-x2+x2-x+1
=x.(x9-1)+x2.(x3-1)+(x2+x+1)
=x.(x3-1)(x3+1)+x2(x3-1)+(x2-x+1)
=x.(x-1)(x2+x+1)(x3+1)+x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x.(x-1)(x3+1)+x2(x-1)+1]
=(x2+x+1)(x5+x2-x4-x+x3-x2+1)
=(x2+x+1)(x5-x4+x3-x+1)
Trả lời:
\(x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
...
\(x^{10}-1=\left(x^5-1\right)\left(x^5+1\right)\)
\(=\left(x-1\right)\left(x^4+x^3+x^2+1\right)\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)\)