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

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

x³ + 2x² + 2x + 1 

= (x³ + 1) + (2x² + 2x)

= (x + 1)(x² + x + 1) + 2x(x + 1)

= (x + 1)(x² + x + 1 + 2x)

= (x + 1)(x² + 3x + 1)

 Chúc bạn học tốt

Ta có: \(x^3+2x^2+2x+1\)

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

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

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

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

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

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

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

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

`=(-x-2)^2`

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

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

Đặt 2x +1 = t, thay vào ta được:

t⁴ - 3t² +2 

= t⁴ - t² - 2x² + 2

= t²(t² -1) - 2(t² -1)

= (t²-1)(t²-2)

=(t-1)(t+1)(t² -2)

Thay t = 2x+1 vào ta được:

(2x+1-1)(2x+1+1)[(2x+1)² -2]

= 2x . (2x+2) (4x² + 4x+1 -2)

= 4x(x+1)(4x² + 4x - 1) 

(2x-1)³-(2x-1)=(2x-1) [( 2x-1)² -1 ]

                      = 2x (2x -1 ) ( 2x-2)

(2x - 1)³ - (2x - 1)

= (2x - 1)[(2x - 1)² - 1]

= (2x - 1)(4x² - 4x + 1 - 1)

= (2x - 1)(4x² - 4x)

Trả lời:

( 2x - 1 )³ - ( 2x - 1 )

= ( 2x - 1 ) [ ( 2x - 1 )² - 1 ]

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

= 2x ( 2x - 1 ) ( 2x - 2 ) 

1) \(x\left(4x+1\right)\)

2) \(3\left(x-3y\right)\)

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

\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)

= 1 - 4x² - ( x² - 4)

= 1 - 4x² - x² + 4

= -5x² + 5

= - ( 5x² - 5 )

= - 5 (x² - 1)