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

\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

(x² + 2.x.3 + 3² - 1).(x² + 2.x.4 + 16 - 1) - 24

=[(x+3)² - 1]. [(x+4)²-1] -24

=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24

=(x+4)(x+2)(x+5)(x-3) - 24

(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)

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

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

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

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

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

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

7x - 6x² -2 = -6x²  +3x + 4x -2= -3x( 2x -1) + 2 ( 2x-1) = (2x-1)(2-3x)

\(x^3-3x^2+3x-1\) =0

=>\(\left(x-1\right)^3\)=0

=>x-1=0

=>x=1

vậy x =1

\(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)