\(4x^2+8xy-3x-6y=4x\left(x+2y\right)-3\left(x+2y\right)=\left(4x-3\right)\left(x+2y\right)\)

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

\(x^3-5x^2-14x=x\left(x^2-5x-14\right)=x\left(x^2-7x+2x-14\right)=x\left[x\left(x-7\right)+2\left(x-7\right)\right]=x\left(x-7\right)\left(x+2\right)\)

\(x^4+4y^4=\left(x^2\right)^2+2\times x^2\times2y^2+\left(2y^2\right)^2-4x^2y^2=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)

^2 + 4xy - 16 + 4y^2 
= x^2 + 4xy + 4y^2 - 4^2 
= (x + 2y)^2 - 4^2 
= (x + 2y - 4)(x + 2y + 4) 
2x^2-5xy-3y^2 
= 2^x + xy - 6xy - 3y^2 
= x(2x + y) - 3y(2x + y) 
= (2x + y)(x - 3y)

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Mày ra câu hỏi từ từ người ta trả lới cho chứ cứ hối người ta 😡

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

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

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

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

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

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

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

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Chon D

D

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 )

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

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

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

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

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

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

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

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

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