Question

[Nương Mạnh]

Thức hiện phép chia (x³-3x²+3x-1)(2y²+4y+2) : (x-1)(y+1)²

Answer 1

[Nguyễn Minh Quang]

ta chú ý hai hằng đẳng thức sau

\(x^3-3x^2+3x-1=\left(x-1\right)^3\)

\(y^2+2y+1=\left(y+1\right)^2\)

từ đây ta sẽ có \(\left(x^3-3x^2+3x-1\right).\left(2y^2+4y+2\right)=\left(x-1\right)^3.2\left(y+1\right)^2\)

vậy \(\left(x^3-3x^2+3x-1\right).\left(2y^2+4y+2\right):\left(x-1\right)\left(y+1\right)^2\)

\(=\left(x-1\right)^3.2\left(y+1\right)^2:\left(x-1\right)\left(y+1\right)^2=2\left(x-1\right)^2\)