$\left(x^2-2\right)\left(x^2-2x+2\right)\left(x^2+2x+2\right)\left(x^2+2\right)$

$=\left[\left(x^2-2\right)\left(x^2+2\right)\right].\left\{\left[\left(x^2+2\right)-2x\right].\left[\left(x^2+2\right)+2x\right]\right\}$

$=\left(x^4-4\right).\left[{\left(x^2+2\right)}^2-4x^2\right]$

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

$=x^8-1$

Trả lời:

( x² - 2x + 2 ) ( x² - 2 ) ( x² + 2x + 2 ) ( x² + 2 )

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

= [ ( x² + 2 )² - ( 2x )² ] ( x⁴ - 4 )

= ( x⁴ + 4x² + 4 - 4x² ) ( x⁴ - 4 )

= ( x⁴ + 4 ) ( x⁴ - 4 )

= ( x⁴ )² - 4²

= x⁸ - 16

(x + 1)² - (2x - 1)² = 0

<=> (x + 1 + 2x - 1) (x + 1 - 2x + 1) = 0

<=> 3x (- x + 2) = 0

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\-x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy tập nghiệm pt: S = {0 ; 2}.

( x + 1 )² - ( 2x - 1 )² = 0

=> ( x + 1 )² = ( 2x - 1 )²

=> x + 1 = 2x - 1

=> x + 2 = 2x

=> 2x - x = 2

=> x = 2

Vậy x = 2

\(m,x^3+48x=12x^2+64\)

\(x^3+48x-12x^2-64=0\)

\(\left(x-4\right)^3=0\)

\(x=4\)

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

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

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

\(x=1\)

\(\Leftrightarrow x^3+48x-12x^2-64=0\)0

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)-12x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(x-4\right)^3=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Trả lời:

j, ( x + 1 )² - ( 2x - 1 )² = 0

<=> ( x + 1 - 2x + 1 ) ( x + 1 + 2x - 1 ) = 0

<=> ( 2 - x ) 3x = 0 

<=> 2 - x = 0 hoặc 3x = 0

<=> x = 2 hoặc x = 0

Vậy x = 2; x = 0 là nghiệm của pt.

k, Sửa đề: 8x³ + 12x - 1 = 6x²

<=> 8x³ + 12x - 1 - 6x² = 0

<=> ( 2x )² - 3.x².2 + 3.x.2² - 1³ = 0

<=> ( 2x - 1 )³ = 0

<=> 2x - 1 = 0

<=> 2x = 1

<=> x = 1/2

Vậy x = 1/2 là nghiệm của pt.

l, x³ + 15x² + 75x + 125 = 0

<=> x³ + 3.x².5 + 3.x.5² + 5³ = 0

<=> ( x + 5 )³ = 0

<=> x + 5 = 0

<=> x = - 5

Vậy x = - 5 là nghiệm của pt.

a) \(x^2-\frac{1}{49}=0\)

<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)

Vậy x = \(\pm\frac{1}{7}\)

b) \(64-\frac{1}{4}x^2=0\)

<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)

<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)

Vậy \(x=\pm16\)

c) 9x² + 12x + 4 = 0

<=> (3x + 2)² = 0

<=> 3x + 2 = 0 

<=> x = -2/3

Vậy x = -2/3

e) \(x^2+\frac{1}{4}=x\) 

<=> \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)

Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng 

\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

`P=27-27x+9x^2-x^3=3^3-3.3^2 .x+3.3.x^2-x^3=(3-x)^3`

Thay `x=-17`: `(3+17)^3=20^3=8000`

`Q=x^3+3x^2+3x=x(x^2+3x+3)`

Thay `x=99`: `Q=99 . (99^2 +3.99+3)=99. 10101=999 999`

`N=(3x-1)^2-2(9x^2-1)+(3x+1)^2

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

`=(3x-1-3x-1)^2=(-2)^2=4`

`x^3-6x^2y+12xy^2-8y^3`

`=x^3-3.x^2 .2y +3.x.(2y)^2-(2y)^3`

`=(x-2y)^3`

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

\(=3x^2\left(3x-1\right)-\left(9x^2-6x+1\right)\)

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

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