Câu hỏi của Trịnh Công Mạnh Đồng

Question

Phân tích đa thức thành nhân tử: $x^3-12x-y^3+6y^2-16=0$

Jup e vs ạ @Akai Haruma; @Aki Tsuki; @Mysterious Person.

Mysterious Person · Accepted Answer

ta có : $x^3-12x-y^3+6y^2-16$

$=x^3-\left(y^3-6y^2+12y-8\right)-12x+12y-24$

$=x^3-\left(y-2\right)^3-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+x\left(y-2\right)+\left(y-2\right)^2\right)-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+y^2+xy-2x-4y-8\right)$Câu trả lời: ta có : $x^3-12x-y^3+6y^2-16$

$=x^3-\left(y^3-6y^2+12y-8\right)-12x+12y-24$

$=x^3-\left(y-2\right)^3-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+x\left(y-2\right)+\left(y-2\right)^2\right)-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)-12\left(x-y+2\right)$

$=\left(x-y+2\right)\left(x^2+y^2+xy-2x-4y-8\right)$

Violympic toán 9