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

    x³ - x² + 3x - 27.

= (x³ - 27) - (x² - 3x).

= (x - 3) (x² + 3x + 9) - x (x - 3).

= (x - 3) (x² + 3x + 9 - x).

= (x - 3) (x² + 2x + 9).

x³ - x² + 3x - 27 = ( x³ - 27 ) - ( x² - 3x )

= ( x - 3 ) ( x² + 3x + 9 ) - x ( x - 3 ) 

= ( x - 3 ) ( x² + 3x + 9 - x )

= ( x - 3 ) ( x² - 2x + 9 )

* Chúc bạn học tốt

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)

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

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

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

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

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

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

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

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

\(x^5-3x^4+3x^3-x^2\)

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

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

hk tốt

3x(x-1)²-(1-x)³=3x(x-1)²-(1-x)².(1-x)=3x(x-1)²+(x-1)²​.(1-x)=(x-1)²(3x-1-x)=(2x-1)(x-1)²

2(x+3)-x³-3x

\(=-x^3-3x+2x+6\)

\(=-x^3-x+6\)

Đa thức này ko phân tích được nha bạn

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

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

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

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

\(=\left(x-1\right)\left[\left(x^2+x+1\right)+3x\right]=\left(x-1\right)\left(x^2+4x+1\right)\)

x^3-3x^2+3x-1 = (x^3+1)-(3x^2-3x)

(Mình sẽ có hằng đẳng thức x^3+1 cũng giống như x^3+1^3 vì 1^3=1 nhé ) 

= ( x^3+1^3)- (3x^2-3x )

=(x-1)* (x^2+ x*1 + 1^2) -( 3x^2-3x)( Chuyển sang hằng đăng thức ) 

=(x-1 ) *(x^2+ x + 1 ) - 3x(x+1)

=(x-1)*(x^2+x+1-3x ) 

 CÓ MỘT BƯỚC LÀ VÌ DẤU TRỪ Ở TRƯỚC NÊN ĐỔI X+1 THÀNH X-1 NHÉ 

Nếu đúng k dùm minha j , cảm ơn

\(x^3+3x^2+3x+2\)

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

\(x^3+3x^2+3x+2\)

\(=x^3+2x^2+x^2+2x+x+2\)

\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)

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