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

Lời giải:

$x^3+y^3+3xy(x+y)=x^3+3x^2y+3xy^2+y^3=(x+y)^3$

P.s: Lần sau bạn chú ý ghi đầy đủ yêu cầu đề nhé.

3x² - x - 3xy + y

= 3x(x-y) - (x-y)

= (3x-1)(x-y)

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

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

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

Bài 1: 

\(=3x^3y-6x^2y^2+15xy\)

Bài 2: 

\(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

\(x^2+2xy-25+y^2\\ =\left(x^2+2xy+y^2\right)-5^2\\ =\left(x+y\right)^2-5^2\\ =\left(x+y-5\right)\left(x+y+5\right)\)

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

Còn 1 câu bên dưới nữa b

a: =x^2(x^2+2x+1)

=x^2(x+1)^2

b: =x^3+3x^2y+3xy^2+y^3-x-y

=(x+y)^3-(x+y)

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

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

c: =5(x^2-2xy+y^2-4z^2)

=5(x-y-2z)(x-y+2z)

\(x^4+1\)

\(=x^4+2x^2+1-2x^2\)

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

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

______

\(4x^4y^4+1\)

\(=4x^4y^4+4x^2y^2+1-4x^2y^2\)

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

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

______

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

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

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

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

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

______

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

\(=x^2+xy+2xy+2y^2\)

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

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

x³ - x + 3x²y + 3xy² + y³ - y ( sửa -x³ -> x³ )

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= ( x + y )³ - ( x + y )

= ( x + y )[ ( x + y )² - 1 ]

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