a) A = [(2x + y) - (2x - y)] . [(2x +y) + (2x - y)]

b) B = [(x - 2y) - 2y]²

\(a,A=\left(2x+y\right)^2-\left(2x-y\right)^2\\ =\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\\ =2y\cdot4x\\ =8xy\\ b,B=\left(x-2y\right)^2-4y\left(x-2y\right)+4y^2\\ =x^2-4xy+4y^2-4xy+8y^2+4y^2\\ =x^2+16y^2-8xy\\ =\left(x-4y\right)^2\)

\(a,A=\left(2x+y\right)^2-\left(2x-y\right)^2\)

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

\(=2y.4x=8xy\)

Vậy \(A=8xy\)

\(----------\)

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

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

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

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

Vậy \(B=\left(x-4y\right)^2\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

( x + 2y )² - ( 2x - y ) ( x + 2y ) 

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

= ( x + 2y ) ( 3y - x )

Tại x = 3 

=> ( 3 + 2y ) ( 3y - 3 )

Theo mik là vậy nha :D

Hình như bạn thiếu y nha bn

a: \(=4x^2-2xy+2xy-4y^2=4x^2-4y^2\)

b: \(=x^n+xy^{n-1}-y^n-x+y\)

\(\dfrac{x+y}{2\left(x+y\right)}=\dfrac{0}{2.0}=\dfrac{0}{0}???\)

\(A=\dfrac{x+y}{2\left(x+y\right)}\left(đk:x+y\ne0\right)\)

Vậy với \(x+y=0\) thì \(A\in\varnothing\)

Ta có: