a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

b.  \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right).\)

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

vậy................

\(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(x-2\right)\)

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

Vậy ........

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

a) x³ - 7x + 6

= x³ - 2x² + 2x² - 4x - 3x + 6

= x² ( x - 2 ) + 2x ( x - 2 ) - 3 ( x - 2 )

= ( x - 2 ) ( x² + 2x - 3 )

= ( x - 2 ) ( x² - x + 3x - 3 )

= ( x - 2 ) [ x ( x - 1 ) + 3 ( x - 1 ) ] 

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

b ) x³ - 9x² + 6x + 16

= x³ - 8x² - x² + 8x - 2x + 16

= x² ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )

= ( x - 8 ) ( x² - x - 2 )

= ( x - 8 ) ( x² + x - 2x - 2 )

= ( x - 8 ) [ x ( x + 1 ) - 2 ( x + 1 ) ]

= ( x - 8 ) ( x + 1 ) ( x - 2 )

c ) x³ - 6x² - x + 30

= x³ - 5x² - x² + 5x - 6x + 30

= x² ( x - 5 ) - x ( x - 5 ) - 6 ( x - 5 )

= ( x - 5 ) ( x² - x - 6 )

= ( x - 5 ) ( x² - 3x + 2x - 6 )

= ( x - 5 ) [ x ( x - 3 ) + 2 ( x - 3 ) ]

= ( x - 5 ) ( x - 3 ) ( x + 2 )

d ) 2x³ - x² + 5x + 3

= 2x³ + x² - 2x² - x + 6x + 3

= x² ( 2x + 1 ) - x ( 2x + 1 ) + 3 ( 2x + 1 )

= ( 2x + 1 ) ( x² - x + 3 )

a)  \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)

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

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

b)  \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)

\(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)

a, = (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4) = (x-1).(x-2)^2

b, = (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)

k mk nha

a)= (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4)

    = (x-1).(x-2)^2

b)= (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ]

 = (x+1).(x-2).(x-8)

P/s tham khảo nha

5x^2+10xy+5y^2

=5.(x²+2xy+y²)

=5.(x+y)²

x^3-6x^2+9x

=x.(x²-6x+9)

=x.(x-3)²

xy+y^2-x-y

=y.(x+y)-(x+y)

=(x+y)(y-1)

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