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

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

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

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

\(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{\left(x^2+2xy+y^2\right)+xy+y^2}{\left(x^3+x^2y+xy^2+y^3\right)+x^2y-2xy^2-3y^3}\)

\(=\frac{\left(x+y\right)^2+y\left(x+y\right)}{\left(x+y\right)^3+y.\left(x^2-2xy-2y^2\right)}\)

( 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

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

\(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\dfrac{\left(x+y\right)\left(x+2y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)

\(=\dfrac{x+y}{x^2-y^2}\)

\(=\dfrac{1}{x-y}\)