x\(^2\)-(a+b)x+ab

= x\(^2\)-ax-bx+ab

= x(x-a) - b(x-a)

= ( x-a).( x-b)

ax-2x-a\(^2\)+2a

= x(a-2) - a(a-2)

= (a-2).( x-a)

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

#) TL :

x³ - 2x - 4

= x³ - 4x + 2x - 4

= x( x² - 4 ) + 2( x - 2)

= x( x -2 )( x + 2)  + 2(x-2)

= (x- 2)( x² + 2x + 2 )

Chúc bn hok tốt ạ :3

Cách 1:  Như bạn kia

Cách 2: Muốn thêm bớt thì thêm bớt:)

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

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

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

Cách 3: Tách hạng tử:

\(x^3-2x-4=\left(x^3-8\right)-\left(2x-4\right)\)

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

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

Cách 4: Tách hạng tử:

\(x^3-2x-4=\frac{1}{2}x^3-2x+\frac{1}{2}x^3-4\)

\(=\frac{1}{2}x\left(x^2-4\right)+\frac{1}{2}\left(x^3-8\right)\)

Dùng hằng đẳng thức tiếp xem có ra không:D

\(x^3+x^2+4\)

\(=x^3-x^2+2x^2+2x-2x+4\)

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

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

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

x³ + x² + 4 

= x³+ x² + 4 + 4³ - 4³

= (x + 4)³ - 4³

⁼ [(x+ 4 - 4)] [(x+4)²+ (x+4).4 + 4²] 

#) TL :
x³ + x² + 4

= x³ - x² + 2x² + 2x - 2x + 4

= x( x² - x + 2 ) + 2( x² - x + 2 )

= ( x + 2 )( x²- x + 2 )

Chúc bn hok tốt ạ ;3

x²-2x-3 = x²-3x+x-3 = x(x-3) + (x-3) = (x+1)(x-3)

CHọn mình nha :)

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

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

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

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

hk tot

^^

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

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

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

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

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)

=(x+1)(x^4+x^3+x^2+x+1)