Phân tích đa thức thành nhân tử
x^8+x^7+1 bằng phuong pháp thêm bớt hạng tử x^2+x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
\(x^3+x^2+4\)
\(=\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(2x+4\right)\)
\(=x^2.\left(x+2\right)-x.\left(x+2\right)+2.\left(x+2\right)\)
\(=\left(x+2\right).\left(x^2-x+2\right)\)
\(x^8+x^4+1\)
\(=x^4.\left(x^4+1\right)+\left(x^4+1\right)-x^4\)
\(=\left(x^4+1\right).\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right).\left(x^4+1+x^2\right)\)
\(x^4+2x^2-24\)
\(=x^4-4x^2+6x^2-24\)
\(=x^2\left(x^2-4\right)+6\left(x^2-4\right)\)
\(=\left(x^2+6\right)\left(x^2-4\right)\)
\(=\left(x^2+6\right)\left(x-2\right)\left(x+2\right)\)
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
\(x^4+1\)
\(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-2x^2\)
\(=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)\)
\(x^8+x^7+1\)
\(=x^8-x^2+x^7-x+x^2+x+1\)
\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+x^2+x+1\)
\(=\left(x^2+x\right)\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x\right)\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x\right)\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^5+x^4+x^2+x\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^6-x^4+x^3-x\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^6-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)
Chúc bạn học tốt.