Câu hỏi của Ngọc Linhh

Question

Phân tích đa thức thành nhân tử:

(x+y)(x+2y)(x+3y)(x+4y)+y4

Sáng · Accepted Answer

$\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4$

$=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4$

$=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4$

$=\left(x^2+5xy+5y^2-y^2\right)\left(x^2+5xy+5y^2+y^2\right)+y^4$

$=\left(x+5xy+5y^2\right)\left(x^2+5xy+5y^2\right)-y^4+y^4$

$=\left(x^2+5xy+5y^2\right)^2$Câu trả lời: $\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4$

$=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4$

$=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4$

$=\left(x^2+5xy+5y^2-y^2\right)\left(x^2+5xy+5y^2+y^2\right)+y^4$

$=\left(x+5xy+5y^2\right)\left(x^2+5xy+5y^2\right)-y^4+y^4$

$=\left(x^2+5xy+5y^2\right)^2$

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