đặt 3 xy làm chung nha bn ...
~ hok tốt ~
\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)
\(=3xy\left(x^2-2x-y^2-2yz-x^2+1\right)\)
\(=3xy\left(\left(x^2-2x+1\right)-\left(y^2+2yz+z^2\right)\right)\)
\(=3xy\left(\left(x-1\right)^2-\left(y-z\right)^2\right)\)
\(=3xy\left(x-1+y-z\right)\left(x-1-y+z\right)\)
Đề = 3xy ( x2 - 2x - y2 - 2yz - z2 +1 )
= 3xy [ ( x2 - 2x +1) - ( y2 + 2yz + z2 )]
= 3xy [ ( x- 1 )2 - ( y +z )2
= 3xy ( x - 1 +y + z ) ( x - 1 - y -z)
\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)
\(=\)\(3xy\left(x^2-2x-y^2-2yz-z^2+1\right)\)
\(=\)\(3xy\left[\left(x^2-2x+1\right)-\left(y^2+2yz+z^2\right)\right]\)
\(=\)\(3xy\left[\left(x-1\right)^2-\left(y+z\right)^2\right]\)
\(=\)\(3xy\left(x+y+z-1\right)\left(x-y-z-1\right)\)
Chúc bạn học tốt ~
Phân tích thành nhân tử:
\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)
\(=3xy\left(x^2-2x-y^2-2yz-z+1\right)\)
Bài nếu có nhầm lẫn mong các bạn sửa giùm ah !
