\(x^3-25x=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)

\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)

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

16) 2x + 2y - x² - xy = ( 2x + 2y ) - ( x² + xy ) = 2( x + y ) - x( x + y ) = ( x + y )( 2 - x )

17) x² - 2x - 4y² - 4y = ( x² - 4y² ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2( x + 2y ) = ( x + 2y )( x - 2y - 2 )

18) x²y - x³ - 9y + 9x = ( x²y - x³ ) - ( 9y - 9x ) = x²( y - x ) - 9( y - x ) = ( y - x )( x² - 9 ) = ( y - x )( x - 3 )( x + 3 )

19) x²( x - 1 ) + 16( 1 - x ) = x²( x - 1 ) - 16( x - 1 ) = ( x - 1 )( x² - 16 ) = ( x - 1 )( x - 4 )( x + 4 )

20) 2x² + 3x - 2xy - 3y = ( 2x² - 2xy ) + ( 3x - 3y ) = 2x( x - y ) + 3( x - y ) = ( x - y )( 2x + 3 )

20, \(2x^2+3x-2xy-3y=2x\left(x-y\right)+3\left(x-y\right)=\left(2x+3\right)\left(x-y\right)\)

16, \(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)

17, \(x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x-2y-2\right)\left(x+2y\right)\)

18, \(x^2y-x^3-9y+9x=-x\left(x^2-9\right)+y\left(x^2-9\right)=\left(-x-y\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)

19, \(x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)

Bài 1: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

16) 2x + 2y - x² - xy

   =  ( 2x - x² ) + ( 2y - xy )

   =   x ( 2 - x )  +  y ( 2 - x ) 

   =   ( 2 - x ) ( x + y )

17) x² - 2x - 4y² - 4y

    = ( x² - 4y² ) - ( 2x + 4y )

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

    = ( x + 2y ) ( x - 2y - 2 )

18) x²y - x³ - 9y +9x

    = ( 9x + x³ ) + ( x²y - 9y )

    =  x ( 9 + x² ) +  y ( x² - 9 ) 

    =   x ( 9 + x² ) - y ( 9 + x² )

    = ( 9 + x² ) ( x - y )

    = ( 3 - x ) ( 3 + x ) ( x - y )

19) x² ( x - 1) + 16 (1 - x )

 =   x² ( x - 1 ) - 16 ( x - 1 )

 = ( x - 1 ) ( x² - 16 )

 = ( x - 1 ) ( x - 4 ) ( x + 4 )

20) 2x² + 3x - 2xy - 3y

   =  2x² + 3x - ( 2xy + 3y )

   =   x ( 2x + 3 ) - y ( 2x + 3 )

   = ( 2x + 3 )  ( x - y )

(3x-1)^2 - 16 = (3x-1)^2 - 4^2

                   = (3x-1-4)(3x-1+4)

                   = (3x-5)(3x+3)

\(\left(3x-1\right)^2-16\)

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

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

\(=\left(3x-5\right)\left(3x+3\right)\)

\(=3\left(x+1\right)\left(3x-5\right)\)

ai giải gimf nha

phần này học từ kì 1 rồi mà bn

co cau hoi nao dao ma bao giai

2x - 2y - x² + 2xy - y²

= (2x - 2y) - (x² - 2xy + y²)

= 2(x - y) - (x - y)²

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

c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

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

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

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

c: \(5x^2+15x+3y+xy\)

\(=5x\left(x+3\right)+y\left(x+3\right)\)

\(=\left(x+3\right)\left(5x+y\right)\)

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

\(=\left(x-y\right)^2-9\)

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

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

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)