x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

a: \(5x-20y=5\left(x-4y\right)\)

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

c: \(x\left(x+y\right)-\left(5x+5y\right)=\left(x+y\right)\left(x-5\right)\)

d: \(5\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(y+5\right)\)

\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)

\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)

( x + y + z )³ - x³ - y³ - z³

= [ ( x + y + z )³ - x³ ] - ( y³ + z³ )

= ( x + y + z - x )[ ( x + y + z )² + ( x + y + z )x + x² ] - ( y + z )( y² - yz + z² )

= ( y + z )( 3x² + y² + z² + 2yz + 3zx + 3xy ) - ( y + z )( y² - yz + z² )

= ( y + z )( 3x² + y² + z² + 2yz + 3zx + 3xy - y² + yz - z² )

= ( y + z )( 3x² + 3yz + 3zx + 3xy )

= 3( y + z )( x² + yz + zx + xy )

= 3( y + z )[ ( x² + zx ) + ( xy + yz ) ]

= 3( y + z )[ x( x + z ) + y( x + z ) ]

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

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

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

\(=\left(x+y+z-x\right).\left[\left(x+y+z\right)^2+\left(x+y+z\right).x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right).\left[x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-\left(y^2-yz+z^2\right)\right]\)

\(=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-y^2+yz-z^2\right)\)

\(=\left(y+z\right).\left(3x^2+3xy+3yz+3xz\right)\)

\(=\left(y+z\right).\left[\left(3x^2+3xy\right)+\left(3yz+3xz\right)\right]\)

\(=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]\)

\(=\left(y+z\right).\left(x+y\right).\left(3x+3z\right)\)

\(=3.\left(y+z\right).\left(x+y\right).\left(x+z\right)\)

1D  2C

Câu 1: D

Câu 2: C

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

Làm thế này nek bạn=

[4x (x+y+z)] [(x+y) (x+z)]+(yz)^2=4(x²+yx+xz)(x²+xz+yx+yz)+(yz)^2

Đặt x²+yx+zx=a ta có:

4a(a-yz)+(yz)²=4a²-4ayz+(yz)²=(2a-yz)²( Giờ thì thay a vào nữa là xong ko hỉu đoạn nào cứ ns nha bạn :D