a: =x^2(x^2+2x+1)

=x^2(x+1)^2

b: =x^3+3x^2y+3xy^2+y^3-x-y

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

=(x+y)[(x+y)^2-1]

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

c: =5(x^2-2xy+y^2-4z^2)

=5(x-y-2z)(x-y+2z)

a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

= 5 ( x - y - 2z ) ( x - y + 2z )

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

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

a) \(x^4+2x^3+x^2=\left(x^2\right)^2+2.x^2.x+x^2=\left(x^2+x\right)^2\)

b) \(x^3-x+3x^2y+3xy^2+y^3-y=x^3+3x^2y+3xy^2+y^3-x-y\)

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

c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5}x-\sqrt{5}y\right)^2-20z^2\)

Câu b :

\(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

Câu c :

\(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2\right)-20z^2\)

\(=5\left(x-y\right)^2-20z^2\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

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

a) x⁴ + 2x³ + x² =

= x²(x² + 2x +1)

= x²(x + 1)²

b) x³ - x + 3x² y + 3x y² + y³ - y =

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

= ( x + y)³ - ( x + y)

= ( x + y) [ ( x + y )² - 1]

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

c) 5x² - 10 xy + 5y² - 20z² =

= 5( x² - 2 xy + y² - 4 z² )

= 5[ ( x - y )² - ( 2 z )² ]

= 5( x - y - 2z )( x - y + 2z)

Nhiều thế                  

mình làm bài 1 nhé 

kq sau khi thu gọn là A =x^6 -7+3x

Trời ơi!Làm sao bây giờ,tốt nhất là nên lờ câu hỏi của bạn ấy đi các bạn ạ!

\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)

\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

a) \(x^2-x-y^2-y\)

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

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

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

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

b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)

c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)

                                                   \(=-5\left(4z^2+2xy-y-x^2\right)\)

d)\(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

\(x^2-x-y^2-y\)

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

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

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

hk tốt

a.\(x^2\left(x^2+2x+1\right)\)

   \(x^2\left(x+1\right)^2\)

1/a ) = (x+y)³ -(x+y)

= (x+y)[(x+y)²+1]

c) = 5(x²-xy+y²)-20z²

=5(x-y)²-20z²

= 5 [ (x-y)²- 4z² ]

=5(x-y-4z)(x-y+4z)
 

Bài 1:

a) x³-x+3x²y+3xy²+y³-y

=x³+2x²y-x²+xy²-xy+x²y+2xy²-xy+y³-y²+x²+2xy-x+y²-y

=x(x²+2xy-x+y²-y)+y(x²+2xy-x+y²-y)+(x²+2xy-x+y²-y)

=(x²+2xy-x+y²-y)(x+y+1)

=[x(x+y-1)+y(x+y-1)](x+y+1)

=(x+y-1)(x+y)(x+y+1) 

c) 5x²-10xy+5y²-20z²

=-5(2xy-y²+4z²-2)

Bài 2:

5x(x-1)=x-1   

=>5x²-6x+1=0

=>5x²-x-5x+1

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

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

=>x=1 hoặc x=1/5

b) 2(x+5)-x²-5x=0

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

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

=>x=2 hoặc x=-5