x² + 2xy - 8y² + 2xz + 14yz - 3z²

= ( x² + y² +z² + 2xy + 2yz ) + ( -9x² + 12yz - 4x² )

= ( x + y +z )² - [ (3x)2 - 2.3.x.2y + ( 2x)²

= ( x + y +z )² -  ( 3y - 2x)²

= ( x + y +z - 3y + 2x )(x+ y + z + 3y - 2x )

dảk burh bủh

\(x^2+2xy-8y^2+2xz+14yz-3z^2\)

\(=\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(-9x^2+12yz-4x^2\right)\)

\(=\left(x+y+z\right)^2-\left[\left(3x\right)^2-2.3x.2y+\left(2x\right)^2\right]\).

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

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

anh đi anh nhớ quê nha 

nhớ canh rau muống nhớ cà dầm tương 

nhớ thằng đẩy bố xuống mương 

bố mà bắt được bố tương vỡ mồm

x²+2xz+2xy+4yz

= ( x²+2xz ) + ( 2xy + 4yz )

= x( x + 2z ) + 2y( x +2z ) 

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

Trả lời:

\(x^2+2xz+2xy+4yz\)

\(=\left(x^2+2xz\right)+\left(2xy+4yz\right)\)

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

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

\(x^2+y^2+z^2+2xy+2yz+2xz-\left(9y^2-12yz+4z^2\right)\)

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

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

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

\(x^3-8+x^2-2x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\end{matrix}\right.\)

Ko cần "GP", cảm ơn

Ai trả lời mình đi cho đỡ quê 😞💔

a: \(=\left(x+2-y\right)\left(x+2+y\right)\)

c: \(=\left(x-y\right)^2\)