d) <=>x²-5x-x+5=0

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

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

<=>x=5 hoặc x=1

thank nha

\(a,x^3-3x^2+3x-1=0\)

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

\(\Rightarrow x-1=0\Rightarrow x=1\)

\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)

Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)

Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)

Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)

Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:

\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt

Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)

\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)

\(c,x^3+6x^2+12x+8=0\)

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

\(\Leftrightarrow x+2=0\Rightarrow x=-2\)

\(d,x^3-6x^2+12x-8=0\)

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

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(e,8x^3-12x^2+6x-1=0\)

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

\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)

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

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

\(\Rightarrow x+3=0\Rightarrow x=-3\)

a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)

b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)

c. \(8x^3-12x^2+6x-1=0\)

\(\Rightarrow\left(2x-1\right)^3=0\)

\(\Rightarrow x=\frac{1}{2}\)

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

\(\Rightarrow x-3=-2\Leftrightarrow x=1\) vậy \(x=1\)

b) \(64x^3+48x^2+12x+1=0\Leftrightarrow\left(4x+1\right)^3=3^3\)

\(\Rightarrow4x+1=3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

d) \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\) vậy \(x=-2\)

e) \(x^3-6x^2+12x-8=0\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\) vậy \(x=2\)

Bài 1: Tìm x

a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=0\)

\(\Leftrightarrow-12x-24=0\)

\(\Leftrightarrow-12x=24\)

hay x=-2

Vậy: x=-2

b) Ta có: \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)-11\left(x-1\right)\left(x+1\right)-6=0\)

\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11\left(x^2-1\right)-6=0\)

\(\Leftrightarrow11x^2+6x+12-11x^2+11=0\)

\(\Leftrightarrow6x+23=0\)

\(\Leftrightarrow6x=-23\)

hay \(x=-\frac{23}{6}\)

Vậy: \(x=-\frac{23}{6}\)

c) Ta có: \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

hay \(x=\frac{1}{2}\)

Vậy: \(x=\frac{1}{2}\)

d) Ta có: \(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)

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

\(\Leftrightarrow x+3=0\)

hay x=-3

Vậy: x=-3

a) (2x + 1)² - 4(x + 2)² = 9

4x² + 4x + 1 - 4(x² + 4x + 4) = 9

4x² + 4x + 1 - 4x² - 16x - 16 = 9

-12x - 15 = 9

-12x = 9 + 15

-12x = 24

x = 12 : (-2)

x = -2

b) (3x - 1)² + 2(x + 3)² + 11(x + 1)(1 - x) = 6

9x² - 6x + 1 + 2(x² + 6x + 9) - 11(x + 1)(x - 1) = 6

9x² - 6x + 1 + 2x² + 12x + 18 - 11(x² - 1) = 6

9x² - 6x + 1 + 2x² + 12x + 18 - 11x² + 11 = 6

6x + 30 = 6

6x = 6 - 30

6x = -24

x = -24 : 6

x = -4

c) 8x³ - 12x² + 6x - 1 = 0

(2x)³ - 3.(2x)².1 + 3.2x.1² - 1³ = 0

(2x - 1)³ = 0

2x - 1 = 0

2x = 1

x = 1/2

d) x³ + 9x² + 27x + 27 = 0

x³ + 3.x².3 + 3.x.3² + 3³ = 0

(x + 3)³ = 0

x + 3 = 0

x = 0 - 3

x = -3

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

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

c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)