a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

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

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

a, ( x + 1 ) = 0

<=> x = -1

b, x³ - 9x² + 27x - 27 = 0

<=> ( x - 3 )³ = 0 

<=> x - 3 = 0

<=> x = 3

9x²-6x-3=0

=>9x²-9x+3x-3=0

=>(x-1)(9x-3)=0

=>x-1=0 hoặc 9x+3 = 0

=> x=1 hoặc x=-1/3

b. x³+9x²+27x+19=0

   x³+x²+8x²+8x+19x+19=0

(x+1)(x²+8x+19)=0

x+1=0 => x=-1 

x²+8x+19= x²+8x+16+3=(x+4)²+3 lớn hơn hoặc bằng 3., lớn hơn 0 với moị x

a, \(\Rightarrow3\left(3x^2-2x-1\right)=0\)

\(\Rightarrow3x^2-2x-1=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=1\\3x-2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\3x-2=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\Rightarrow x^3+3x^2+6x^2+9x+18x+19=0\)

\(\Rightarrow x^2\left(x+3\right)+3x\left(x+3\right)+18\left(x+3\right)-2=0\)

\(\Rightarrow\left(x+3\right)\left(x^2+3x+18\right)=2\)

Mk k co thoi gian. buoc tiep theo tu lam not nhe

<=>\(9x^2-6x+1-4=0\)

<=>\(\left(3x-1\right)^2-4=0\)

<=>\(\left(3x-1\right)^2=4\)

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

<=>3x-1=2

<=>3x=3

<=>x=1

Vậy x=1

\(a,9x^2-6x-3=0\)

\(\Leftrightarrow9x^2-6x+1-4=0\)

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

\(\Rightarrow3x-1=\pm2\)

\(\hept{\begin{cases}3x-1=2\Rightarrow x=1\\3x-1=-2\Rightarrow x=\frac{-1}{3}\end{cases}}\)

Vậy \(x=1\) hoặc \(x=\frac{-1}{3}\)

\(b,x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^3+9x^2+27x+27-8=0\)

\(\Leftrightarrow\left(x+3\right)^3=8\)

\(\Rightarrow x+3=2\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

\(c,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=\frac{-11}{25}\)

Vậy \(x=\frac{-11}{25}\)

\(9x^2-6x-3=0\)

<=> \(\left(3x\right)^2-2.3x.1+1-4=0\)

<=> \(\left(3x-1\right)^2-2^2=0\)

<=> \(\left(3x-3\right)\left(3x+1\right)=0\)

<=> \(\hept{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

\(x^3+9x^2+27x+19\) \(=0\)

<=>\(x^3+x^2+8x^2+8x+19x+19=0\)

<=> \(x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

<=> \(\left(x^2+8x+19\right)\left(x+1\right)=0\)

mà \(x^2+8x+19>0\)

=> \(x+1=0\)

<=> \(x=-1\)

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

<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)

<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)

<=>  \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)

<=> \(x^3-25x-x^3+2x^2+4x-8=3\)

<=> \(2x^2-21x-8=3\)

<=> \(2x^2-21x-11=0\)

<=> \(2x^2-22x+x-11=0\)

<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)

<=> \(\left(2x+1\right)\left(x-11\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)

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

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  

a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)

c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)

d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)

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

a) \(x^3+9x^2+27x+19=0\)

\(\Rightarrow x^3+x^2+8x^2+8x+19x+19=0\)

\(\Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\)

Mà \(x^2+8x+19=x^2+2.x.4+16+3=\left(x+4\right)^2+3\)

Vì \(\left(x+4\right)^2\ge0\) với mọi x

\(3>0\)

\(\Rightarrow\left(x+4\right)^2+3>0\) với mọi x

=> ( x + 4 )² + 3 vô nghiệm

=> x + 1 = 0

=> x = -1

Vậy x = -1

b) \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)

\(\Rightarrow\left(2x\right)^3+3.\left(2x\right)^2+3.2x+1+x\left(x^2-2^2\right)-9x\left(x^2-4x+4\right)+57=0\)

\(\Rightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)

\(\Rightarrow48x^2-34x+58=0\)

\(\Rightarrow2\left(24x^2-17x+29\right)=0\)

\(\Rightarrow24x^2-17x+29=0\)

... Tới đây mình bí luôn rồi, sorry

Câu a : \(x^3+9x^2+27x+19=0\)

\(\Leftrightarrow\left(x^3+9x^2+27x+27\right)-8=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Leftrightarrow x+1=0\) ( Vì : \(x^2+8x+19>0\))

\(\Leftrightarrow x=-1\)

Vậy \(x=-1\)

Câu b : \(\left(2x+1\right)^3+x\left(x-2\right)\left(x+2\right)-9x\left(x-2\right)^2+57=0\)

\(\Leftrightarrow8x^3+12x^2+6x+1+x^3-4x-9x^3+36x^2-36x+57=0\)

\(\Leftrightarrow48x^2-34x+58=0\)

\(\Rightarrow PTVN\)

Vậy ko có giá trị của x

a)Ta có: x^3 +9x^2 +27x + 19=0

x^3 + 9x^2 + 27x + 27 -8=0

(x+3)^3 -2^3=0

(x+3-2)(x^2 + 6x+9 +4x+12 +4)=0

(x+1)(x^2 +10x+ 25)=0

=> x+1=0 hoặc x^2 +10x+25=0

=> x=-1 hoặc (x+5)^2 =0

=> x=-1 hoặc x+5=0

=> x=-1 hoặc x=-5