(4x²+2xy+y²)(2x-y)-(2x-y)(4x²-2xy+y²)

=(2x³-y³)-(2x+y)(4x²-2xy+y²)

=(2x³-y³)-(2x³+y³)

=2x³-y³-2x³+y³

=0

a) Ta có: \(\left(y+3\right)\left(y^2-3y+9\right)-\left(60-y^3\right)\)

\(=y^3+27-60+y^3\)

\(=2y^3-33\)

b) Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

c) \(x^2+1+2x-y^2\)

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

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

\(=\left(x+1+y\right)\left(x+1-y\right)\)

d) \(x^2+4x-2xy-4y+y^2\)

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

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

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

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

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

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

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

\(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy.\left(x^2-y^2-2y-1\right)\)

\(=2xy.[x^2-\left(y^2+2y+1\right)]\)

\(=2xy.[x^2-\left(y+1\right)^2]\)

\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)

Vậy chọn đáp án A

chọn A

(2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)

= (2x + y)[(2x)2 – 2x.y + y2] – (2x – y)[(2x)2 + 2x.y + y2]

= [(2x)3 + y3] – [(2x)3 – y3]

= (2x)3 + y3 – (2x)3 + y3

= 2y3

a) (x+3)(x^2-3x+9)-(54+x^3)

= x^3- 3x^2+9x+3x^2-9x+27-54-x63

= -27

b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)

= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]

= [(2x)3^3+ y^3] – [(2x)^3 – y^3]

= (2x)^3 + y^3 – (2x)^3 + y^3

= 2y^3

a)(x+3)(X^2-3x+9)-(54+x^3)

= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)

= 27- 54

= -27

b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)

= \((2x)^3\) + \(y^3\)  - [\((2x)^3\) - \(y^3\) ]

= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)

= \(2y^3\)

(4x² + 2xy + y²)(2x - y) - (2x + y)(4x² - 2xy + y²) = 8x³ - y³ - 8x³ - y³ = - 2y³

\(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=\left(2x-y\right)((2x)^2+2xy+y^2-\left(2x+y\right)((2x)^2-2xy+y^2\)

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

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

\(=-2y^3\)