mày viết lại cái đề bài hộ tao cái

a: \(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+9x\left(x+3\right)=0\)

=>(x+3)(x^2+6x+9)=0

=>x=-3

b: \(\Leftrightarrow x^2-2x-x^3+6x^2-4=0\)

=>-x^3+6x^2-2x-4=0

hay \(x\in\left\{5.5;1.14;-0.64\right\}\)

c: =>(3x-1)^3=8

=>3x-1=2

=>3x=3

=>x=1

\(1-27x^3\)

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

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

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

\(=\left(3x+1\right)^3\)

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

=(1-3x)(1+3x+9x\(^2\)

a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

b: =(1-2x)(1+2x)

c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d: =(x+3)^3

e: \(=\left(2x-y\right)^3\)

f: =(x+2y)(x^2-2xy+4y^2)

d) <=>x²-5x-x+5=0

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

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

<=>x=5 hoặc x=1

thank nha

a ) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

b ) \(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)

\(\Leftrightarrow\left(x-3\right)^3=0\)

\(\Leftrightarrow\left(x-3\right)=0\)

\(\Leftrightarrow x=3\)

a) x³ - 6x² + 12x - 8 = 0

   ( x - 2 ) ³                = 0

    x - 2                      = 0

    x                           = 2

b) x³ + 9x² + 27x + 27 = 0

    ( x + 3 )³                    = 0

      x + 3                         = 0

      x                                = -3

lấy máy tính bỏ túi áấy bn  

=> -9x² + 28x - 27 = 0

=> \(\Delta\)' = b'² - ac = 14² - [ (-9) . (-27) ] = -47 < 0  

vì denta nhỏ hơn 0  nên => pt vô nghiệm

a) x³-9x²+27x-27=0

<=>(x-3)³=0

<=>x-3=0

<=>x=3

b) x³-25x=0

<=>x.(x²-25)=0

<=>x.(x-5)(x+5)=0

<=>x=0 hoặc x-5=0 hoặc x+5=0

<=>x=0 hoặc x=5 hoặc x=-5

c)9x²-1=0

<=>(3x-1)(3x+1)=0

<=>3x-1=0 hoặc 3x+1=0

<=>x=1/3 hoặc x=-1/3

a, x^3 - 9x^2 + 27x - 27 = 0 

=> ( x - 3)^3 = 0 

=> x - 3 = 0 

=> x = 3 

b, x^3 - 25x = 0 

=> x(x^2 - 25) = 0 

=> x(x-5)(x + 5) = 0 

=> x =0 hoặc x - 5 = 0 hoặc x + 5 = 0 

=> x= 0 hoặc x =5 hoặc x = -5 

c, 9x^2 -  1 = 0 

 => (3x)^2 - 1^2 = 0 

=> ( 3x- 1)(3x+ 1) = 0 

=> 3x - 1 = 0 hoặc 3x + 1 = 0 

=> x = 1/3 hoặc x = -1/3