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 )

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

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

a.5x²-10xy+5y²-20z²

  =5(x²-2xy+y²-4z²)

  =5[ (x²-2xy+y²)-(2z)² ]

  =5[ (x-y)²-(2z)² ]

  =5(x-y-2z)(x-y+2z)

b.16x-5x²-3

  =15x+x-5x²-3

  =(15x-3)+(x-5x²)

  =3(5x-1)+x(1-5x)

  =3(5x-1)-x(5x-1)

  =(5x-1)(3-x)

c.x²-5x+5y-y²

  =(5y-5x)+(x²-y²)

  =5(y-x)+(x-y)(x+y)

  =5(y-x)-(y-x)(y+x)

  =(y-x)[5-(y+x)]

  =(y-x)(5-y-x)

d.3x²-6xy+3y²-12z²     (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)

=3(x²-2xy+y²-4z²)

=3[ (x²-2xy+y²)-(2z)² ]

=3[ (x-y)²-(2z)² ]

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

e.x²+4x+3

=x²+3x+x+3

=(x²+x)+(3x+3)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f.(x²+1)²-4x²

=(x²+1)²-(2x)²

=(x²+1-2x)(x²+1+2x)

h.x²-4x-5

=x²-5x+x-5

=(x²+x)+(-5x-5)

=x(x+1)-5(x+1)

-(x+1)(x-5)

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)

1)\(x^4+2x^3+x^2\)

=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra

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

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

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

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

3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)

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

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

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

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

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

k chi nha

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, \(14x^2y-21xy^2-28x^2y^2=7xy\left(2x-3y-4xy\right)\)

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,\(10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x-8\right)\)

d,\(\left(3x+1\right)^2-\left(x+1\right)^2=9x^2+6x+1-x^2-2x-1=7x^2-4x=x\left(7x-4\right)\)

e,\(x^3+y^3+z^3-3xyz=\)=a3+3ab(a+b)+b3+c3−3abc−3ab(a+b)
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