a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

Bài 4:

a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)

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

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

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

Thay \(x=6\) vào (1) ta được:

\(\left(6+4\right)^3=10^3=1000\)

Vậy...........

b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)

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

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

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

Thay \(x=22\) vào (2) ta được:

\(\left(22-2\right)^3=20^3=8000\)

Vậy.............

Chúc bạn học tốt!!!

Bài 2:

a, \(\left(x+9\right)^3=27=3^3\)

\(\Rightarrow x+9=3\Rightarrow x=-6\)

Vậy.........

b, \(8-12x-x^3+6x^2=-64\)

\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)

\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)

\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)

\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)

Vậy............

Chúc bạn học tốt!!!

4. Tính giá trị biểu thức

a) x³ + 12x² + 48x + 64 khi x = 6

Ta có:

x³ + 12x² + 48x + 64 =

= (x³ + 64) + (12x² + 48x)

= (x³ + 4³) + 12x(x + 4)

= (x + 4)(x² - 4x + 4²) + 12x(x + 4)

= (x + 4)(x² - 4x + 16 +12x)

= (x + 4)(x² + 8x + 16)

= (x + 4)(x + 4)²

= (x + 4)³

Thế x = 6 vào biểu thức vừa tìm, ta được:

(x + 4)³ = (6 + 4)³ = 10³ = 1000

Vậy 1000 là giá trị của biểu thức x³ + 12x² + 48x + 64 khi x = 6.

b) x³ - 6x² + 12x - 8 khi x = 22

Ta có:

x³ - 6x² + 12x - 8 =

= (x³ - 8) - (6x² - 12x)

= (x³ - 2³) - 6x(x - 2)

= (x - 2)(x² + 2x + 2²) - 6x(x - 2)

= (x - 2)(x² + 2x + 4 - 6x)

= (x - 2)(x² - 4x + 4)

= (x - 2)(x - 2)²

= (x - 2)³

Thế x = 22 vào biểu thức vừa tìm, ta được:

(x - 2)³ = (22 - 2)³ = 20³ = 8000

Vậy 8000 là giá trị của biểu thức x³ - 6x² + 12x - 8 khi x = 22.

1) \(8x^3-12x^2+6x-1=0\)

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

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

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

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

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

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

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

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

Vậy x = 0 là nghiệm của pt

x^3-6x^2+12x-8=(x-2)^3

\(x^3-6x^2+12x-8=x^3+3.2.x^2+3.2^2.x-2^3=\left(x-2\right)^3\)

\(2x-1^3+8\)

\(=2x-9\)

\(=\left(\sqrt{2x}\right)^2-3^2\)

\(=\left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)\)

_________

\(8x^3-12x^2+6x-1\)

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

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

_______________

\(8x^3-12x^2+6x-2\)

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

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

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

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

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

\(=\left(3x-1\right)\left(3x^2-5x+1\right)\)

b: 8x^3-12x^2+6x-1

=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

=2(x-1)(4x^2-4x+1+2x)

=2(x-1)(4x^2-2x+1)

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

\(=\left(x+2\right)^3=\left(3+2\right)^3=5^3=125\)

\(x^3+6x^2+12x+8\\=(x+2)^3\\Thay.x=3.Ta có:\\(3+2)^3=5^3=125\)