a: \(4x^2-4x\)

\(=4x\cdot x-4x\cdot1\)

\(=4x\left(x-1\right)\)

b: \(x^2-2xy+y^2-4\)

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

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

`4x^2-4x`

`= 4x(x-1)`

__

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

`= (x^2-2xy+y^2)-2^2`

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

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

a,

\(4x^2-4x\\=4x(x-1)\)

b,

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

\(\text{#}Toru\)

Đề sai nhé .Sửu lại

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

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

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

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

\(\left(4x^2-4x+1\right)-\left(x-1\right)^2\)

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

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

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

x² - 4x -5

= x² - 5x + x -5

= x(x-5) + (x-5)

=(x+1)(x-5)

x^2-4x-4=x²-2.2.x-2²

           =[x-2]²

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

Đặt: x - y = a ; 3x + y - z = b ; -4x + z = c 

Ta có: a + b +  c  = x - y + 3x + y - z - 4x + z = 0 

Khi đó: \(\left(x-y\right)^3+\left(3x+y-z\right)^3+\left(-4x+z\right)^3\)

= \(a^3+b^3+c^3\)

= \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(0.\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(3abc\)

= \(3\left(x-y\right)\left(3x+y-z\right)\left(-4x+z\right)\)

cảm ơn ạ 

tôi mới học lớp 2