3x³ + 6x²y+ 3xy²

=3x³ + 3x²y + 3x²y + 3xy²

= 3x²(x +y) + 3xy(x +y)

=(x+y)(3x² + 3xy)

= 3x(x+y)(x+y)

= 3x(x+y)²

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

\(=x^2\left(x+9\right)-4\left(x+9\right)\)

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)

\(=3xy\left(x^2-2x-y^2-2yz-z^2+1\right)\)

\(=3xy\left[\left(x-1\right)^2-\left(y+z\right)^2\right]\)

\(=3xy\left(x-1-y-z\right)\left(x-1+y+z\right)\)

a, x2 + 2y2 - 3xy + x - 2y = x2 - 4xy + 4y2 + xy - 2y2 + x - 2y

= ( x - 2y )2 + y( x - 2y ) + ( x - 2y )

= ( x - 2y )( x - 2y + y + 1 )

( x - 2y )( x - y + 1 )

That so easy bài này chệ làm rồi :))))))) mới lớp 7

x4+2x3+3x2+2x+1x4+2x3+3x2+2x+1 

Đặt y=x2+1y=x2+1, ta có : 

A=x4+2x3+3x2+2x+1A=x4+2x3+3x2+2x+1

=x4+2x2+1+x2+2x+2x3=x4+2x2+1+x2+2x+2x3

=(x2+1)2+2x(x2+1)+x2=(x2+1)2+2x(x2+1)+x2 

=y2+2xy+x2=y2+2xy+x2 

=(x+y)2=(x2+x+1)

Mấy bài này that so easy