Câu hỏi của Trần Mỹ Chi

Question

Phân tích đa thức thành nhân tửa)x^5-x^4-1b)x^8+x^7+1

Ngô Chi Lan · Accepted Answer

a) $x^5-x^4-1$$=\left(x^5+x^2\right)-\left(x^4+x\right)-\left(x^2-x+1\right)$$=x^2\left(x^3+1\right)-x\left(x^3+1\right)-\left(x^2-x+1\right)$$=x^2\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)$$=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)$$=\left(x^2-x+1\right)\left(x^3-x-1\right)$Câu trả lời: a) $x^5-x^4-1$$=\left(x^5+x^2\right)-\left(x^4+x\right)-\left(x^2-x+1\right)$$=x^2\left(x^3+1\right)-x\left(x^3+1\right)-\left(x^2-x+1\right)$$=x^2\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)$$=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)$$=\left(x^2-x+1\right)\left(x^3-x-1\right)$

Ngô Chi Lan · Answer

b) $x^8+x^7+1$$=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)$$=x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)$$=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)$$=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]$$=\left(x^2+x+1\right)\left[\left(x^3-x\right)\left(x^3+1\right)+1\right]$$=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)$Câu trả lời: b) $x^8+x^7+1$$=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)$$=x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)$$=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)$$=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)$$=\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]$$=\left(x^2+x+1\right)\left[\left(x^3-x\right)\left(x^3+1\right)+1\right]$$=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)$

Lâm Linh Ngọc · Answer

a) $x^5-x^4-1=x^5+x^2-x^4-x^2-1$$=x^2\left(x^3+1\right)-\left(x^4+x^2+1\right)=x^2\left(x+1\right)\left(x^2-x+1\right)-\left[\left(x^2\right)^2+2x^2+1-x^2\right]$$=x^2\left(x+1\right)\left(x^2-x+1\right)-\left[\left(x^2+1\right)-x^2\right]$$=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\left(x^2+x+1\right)$$=\left(x^2-x+1\right)\left[x^2\left(x+1\right)-\left(x^2+x+1\right)\right]$$=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)$$=\left(x^2-x+1\right)\left(x^3-x-1\right)$b) $x^8+x^7+1=x^8+x^7+x^6-x^6+1$$=x^6\left(x^2+x+1\right)-\left(x^6-1\right)=x^6\left(x^2+x+1\right)-\left[\left(x^3\right)^2-1\right]$$=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)$$=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)$Mong cô Chuy cho e thêm 1 Gp nựa nha cô '-'Câu trả lời: a) $x^5-x^4-1=x^5+x^2-x^4-x^2-1$$=x^2\left(x^3+1\right)-\left(x^4+x^2+1\right)=x^2\left(x+1\right)\left(x^2-x+1\right)-\left[\left(x^2\right)^2+2x^2+1-x^2\right]$$=x^2\left(x+1\right)\left(x^2-x+1\right)-\left[\left(x^2+1\right)-x^2\right]$$=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\left(x^2+x+1\right)$$=\left(x^2-x+1\right)\left[x^2\left(x+1\right)-\left(x^2+x+1\right)\right]$$=\left(x^2-x+1\right)\left(x^3+x^2-x^2-x-1\right)$$=\left(x^2-x+1\right)\left(x^3-x-1\right)$b) $x^8+x^7+1=x^8+x^7+x^6-x^6+1$$=x^6\left(x^2+x+1\right)-\left(x^6-1\right)=x^6\left(x^2+x+1\right)-\left[\left(x^3\right)^2-1\right]$$=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)$$=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)$Mong cô Chuy cho e thêm 1 Gp nựa nha cô '-'

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