Với x = -3 ta có -27-4*9+ 36+27=0 do đó đa thức chứa nhân tử x+3 
Ta có: x^3 -4x^2-12x+27 = x^3 +3x^2 -7x^2-21x+9x+27 =(x^3 +3x^2)-(7x^2+21x) + (9x+27) =x^2(x+3) -7x(x+3)+ 9(x+3)=(x+3)(X^2 - 7x+9) 
* Xét x^2 -7x + 9 = x^2 - 2x.7/2 +49/4-49/4+9 = (x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2) 
Vậy: x^3 -4x^2-12x+27 = (x+3)(x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2)

k cho mình nha

 = (x+2)(x-2) +(x-2)² = (x-2)(x+2 +x-2) = 2x(x-2)

=( x³ +3³) - (4x² + 12x) 

= (x+3)(x² - 3x + 9 -4x)

=(x+3)(x² -7x +9)

Giúp mình vs mình đang gấp

\(=x^3+3x^2-7x^2-21x+9x+27=\left(x+3\right)\left(x^2-7x+9\right)\)

Đáp án:
   x³-4x²-12x+27
= (x+3)(x²-7x+9)

\(a.\) \(ax^2-a^2x-x+a\)

\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(ax-1\right)\left(x-a\right)\)

\(b.\) \(18x^3-12x^2+2x\)

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

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

\(c.\) \(x^3-5x^2-4x+20\)

\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)

\(=x^2\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x^2-4\right)\left(x-5\right)\)

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

\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)

\(=x^2+15x+7x+105+15\)

\(=x^2+22x+120\)

\(=\left(x+10\right)\left(x+12\right)\)

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: \(xy-4y-5x+20\)

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

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

Đề sai rồi bạn

x³-12x-4x²+27

=(x³+27)-(12x+4x²)

=(x+3)(x²-3x+9)-4x(x+3)

=(x+3)(x²-3x+9-4x)

=(x+3)(x²-7x+9)

\(x^3-12x-4x^2+27\)

\(=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

a) x² – 4 + (x – 2)²

= (x² – 2²) + (x – 2)² = (x – 2)(x + 2) + (x – 2)²

= (x – 2) [(x + 2) + (x – 2)]

= (x – 2)(x + 2 + x – 2)

= 2x(x – 2)

b) x³ – 2x² + x – xy²

= x(x² – 2x + 1 – y²) = x[(x² – 2x + 1) – y2]

= x[(x – 1)2 – y2]

= x[(x – 1) + y] [(x – 1) – y]

= x(x – 1 + y)(x – 1 – y)

c) x³ – 4x² – 12x + 27

= (x³ + 27) – 4x(x + 3)

= (x + 3)(x² – 3x + 9) – 4x(x + 3)

= (x + 3)(x² – 3x + 9 – 4x)

= (x + 3)(x² – 7x + 9)

a) $x^2-4+{\left(x-2\right)}^2$

=x²-2²+(x-2)²

= (x-2)(x+2)+(x-2)²

= (x-2)(x+2+x-2)

= (x-2)2x

b) $x^3-2x^2+x-x{y}^2$

= x(x²-2x+1-y²)

= x[(x²-2x+1)-y²]

=x[(x-1)²-y²]

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

c) $x^3-4x^2-12x+27$

$\mathrm{=x3+3x2-7x2-21x+9x+27}$

$\mathrm{=(x3+3x2)-(7x2+21x)+(9x+27)}$

=x²(x+3)-7x(x-3)+9(x+3)

=(x+3)(x2-7x+9)

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

k

cho

mk 

mk 

giải 

cho

\(x^3-4x^2+12x-27=x^3-27-4x^2+12x\)

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

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

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2