Câu hỏi của Thanh

Question

Phân tích đa thức thành nhân tử$x^3-3xy^2-2y^3$

Lê Tài Bảo Châu · Accepted Answer

$x^3-3xy^2-2y^3$$=x^3-xy^2-2xy^2-2y^3$$=x\left(x^2-y^2\right)-2y^2\left(x+y\right)$$=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)$$=\left(x+y\right)\left(x^2-xy-2y^2\right)$$=\left(x+y\right)\left(x^2-2xy+xy-2y^2\right)$$=\left(x+y\right)\left[x\left(x-2y\right)+y\left(x-2y\right)\right]$$=\left(x+y\right)^2\left(x-2y\right)$Câu trả lời: $x^3-3xy^2-2y^3$$=x^3-xy^2-2xy^2-2y^3$$=x\left(x^2-y^2\right)-2y^2\left(x+y\right)$$=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)$$=\left(x+y\right)\left(x^2-xy-2y^2\right)$$=\left(x+y\right)\left(x^2-2xy+xy-2y^2\right)$$=\left(x+y\right)\left[x\left(x-2y\right)+y\left(x-2y\right)\right]$$=\left(x+y\right)^2\left(x-2y\right)$

Minh Nguyen · Answer

$x^3-3xy^2-2y^3$$=x^3-xy^2-2xy^2-2y^3$$=x\left(x^2-y^2\right)-2y^2\left(x+y\right)$$=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)$$=\left(x+y\right)\left[x\left(x+y\right)-2y^2\right]$$=\left(x+y\right)\left(x^2+xy-2y^2\right)$$=\left(x+y\right)\left(x^2+2xy-xy-2y^2\right)$$=\left(x+y\right)\left[x\left(x-2y\right)-y\left(x-2y\right)\right]$$=\left(x-y\right)\left(x-y\right)\left(x-2y\right)$$=\left(x-y\right)^2\left(x-2y\right)$Câu trả lời: $x^3-3xy^2-2y^3$$=x^3-xy^2-2xy^2-2y^3$$=x\left(x^2-y^2\right)-2y^2\left(x+y\right)$$=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)$$=\left(x+y\right)\left[x\left(x+y\right)-2y^2\right]$$=\left(x+y\right)\left(x^2+xy-2y^2\right)$$=\left(x+y\right)\left(x^2+2xy-xy-2y^2\right)$$=\left(x+y\right)\left[x\left(x-2y\right)-y\left(x-2y\right)\right]$$=\left(x-y\right)\left(x-y\right)\left(x-2y\right)$$=\left(x-y\right)^2\left(x-2y\right)$

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