Violympic toán 8

Phân tích đa thức thành nhân tử:

A. x^4+4y^4

B. x^8+x^4+1

C. x^11+x+1

Cần gấp!!!

Trần Thanh Phương 15 tháng 7 2019 lúc 16:22

A. $x^4+4y^4$

$=\left(x^2\right)^2+2\cdot x^2\cdot2y^2+\left(2y^2\right)^{^2}-2\cdot x^2\cdot2y^2$

$=\left(x^2+2y^2\right)^2-\left(2xy\right)^2$

$=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)$

B. $x^8+x^4+1$

$=x^8+2\cdot x^4\cdot1+1-x^4$

$=\left(x^4+1\right)^2-\left(x^2\right)^2$

$=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)$

$=\left(x^4-x^2+1\right)\left(x^4+2x^2+1-x^2\right)$

$=\left(x^4-x^2+1\right)\left[\left(x^2+1\right)^2-x^2\right]$

$=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)$

Bình luận (0)
Trần Thanh Phương 15 tháng 7 2019 lúc 16:30

C. $x^{11}+x+1$

$=x^{11}-x^{10}+x^{10}-x^9+x^9-x^8+x^8-x^7+x^7-x^6+x^6-x^5+x^5-x^4+x^4-x^3+x^3+x^2-x^2+x+1$

$=\left(x^{11}-x^{10}+x^8-x^7+x^5-x^4+x^2\right)+\left(x^{10}-x^9+x^7-x^6+x^4-x^3+x\right)+\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)$

$=x^2\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)+x\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)+\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)$

$=\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\left(x^2+x+1\right)$

Bình luận (0)
tthnew 15 tháng 7 2019 lúc 19:09

Trần Thanh Phương câu c anh dài dòng:v em nhớ ko nổi:v

$x^{11}+x+1=\left(x^{11}+x^{10}+x^9\right)-\left(x^{10}+x^9+x^8\right)+\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)$

$=x^9\left(x^2+x+1\right)-x^8\left(x^2+x+1\right)+x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)$

$=\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\left(x^2+x+1\right)$

Bình luận (2)

Các câu hỏi tương tự