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) `x^4+2x^3-4x-4`

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

`=(x^2-2)(x^2+2)+2x(x^2-2)`

`=(x^2-2)(x^2+2+2x)`

b) `x^3-4x^2+12x-27`

`=(x^3-27)-(4x^2-12x)`

`=(x-3)(x^2+3x+9)-4x(x-3)`

`=(x-3)(x^2+3x+9-4x)`

`=(x-3)(x^2-x+9)`

c) `xy-4y-5x+20`

`=y(x-4)-5(x-4)`

`=(y-5)(x-4)`

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

b) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)

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

c) Ta có: \(xy-4y-5x+20\)

\(=y\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(y-5\right)\)

\(4x^4y-4x^2y^3+12x^3y+12x^2y^2\)

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

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

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

a)=x²-2x+1-y²-2y-1

=(x-1)²-(y+1)²

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

mình cũng vậy

mình cũng vậy

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

a) x^2 - 4 + ( x - 2 )^2 

= ( x- 2 )(x + 2 ) + ( x-  2)^2 

= ( x - 2 ) ( x + 2 + x - 2 )

= 2x (x-2)

b) x^3 - 2x^2 + x - xy^2

= x ( x^2 - 2x + 1 - y^2) 

= x [ ( x - 1 )^2 - y^2 ] 

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

c) x^3 - 4x^2 - 12x + 27 

= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27 

= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )

Để hai lần nha 

= ( x+ 3 )(x^2 - 7x + 9 ) 

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

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

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

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

hk tốt

^^

a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)

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

c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)

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