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23 tháng 12 2023

a: \(x^2-y^2+3x+3y\)

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

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

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

b: Sửa đề: \(x^2-4y^2+4x+4\)

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

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

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

20 tháng 10 2023

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)

23 tháng 5 2022

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

9 tháng 9 2020

           Bài làm :

 \(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

 \(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

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

 \(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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

\(d ) x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

13 tháng 10 2016

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

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

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

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

13 tháng 10 2016

thank bạn nhiều nha

5 tháng 8 2021

f) = x2( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )

g) = 4( x - y ) + ( x - y )2 = ( x - y )( x - y + 4 )

h) = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )

i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )

j) = ( x - y )( x2 + xy + y2 ) - 3( x - y ) = ( x - y )( x2 + xy + y2 - 3 )

5 tháng 8 2021

Trả lời:

f, x3 - 4x2 - 9x + 36 = ( x3 - 4x2 ) - ( 9x - 36 ) = x2 ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x2 - 9 ) = ( x - 4 )( x - 3 )( x + 3 )

g, 4x - 4y + x2 - 2xy + y2 = ( 4x - 4y ) + ( x2 - 2xy + y2 ) = 4 ( x - y ) + ( x - y )2 = ( x - y ) ( 4 + x - y )

h, x4 + x3 + x2 - 1 = ( x4 + x3 ) + ( x2 - 1 ) =  x3 ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 ) 

i, x2 - y2 - 4x - 4y = ( x2 - y2 ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )

j, x3 - y3 - 3x + 3y = ( x3 - y3 ) - ( 3x - 3y ) = ( x - y )( x2 + xy + y2 ) - 3 ( x - y ) = ( x - y )( x2 + xy + y2 - 3 ) 

1 tháng 10 2016

a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

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

c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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