\(x^4-x+2008x^2+2008x+2008\)

\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

a, x^2 + 7x + 6

= x^2 + x + 6x + 6

= x(x + 1) + 6(x + 1)

= (x + 6)(x + 1)

\(x^2+7x+6\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+6\right)\left(x+1\right)\)

\(a,x^2+7x+6\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

\(b,x^4+2008x^2+2007x+2008\)

\(=x^4-x+2008x^2+2008x+2008\)

\(=x\left(x^3-1\right)+2008\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2008\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

x ² - 2007x - 2008

= x ² + x - 2008x - 2008

= ( x ² + x ) - ( 2008x + 2008 )

= x ( x + 1 ) - 2008 . ( x + 1 )

= ( x + 1 ) . ( x - 2008 )

x^2-4+4xy-8y=x^2+4xy+4y^2-4y^2-8y-4=(x+2y)^2-(2y+2)^2=(x+2y-2y+2)(x+2y+2y-2)=(x+2)(x+4y-2)

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

Thay `x = 2` ta được :

`x^4+x^3-9x^2+10x-8`

`= 2^4 + 2^3 - 9*2^2 + 10*2 - 8`

`= 16 + 8 - 36 + 20 - 8`

`= 0`

Vậy `x = 2` là nghiệm của phương trình trên 

Do đó ta thực hiện phép chia :

\(\left(x^4+x^3-9x^2+10x-8\right):\left(x-2\right)\)

Vậy \(x^4+x^3-9x^2+10x-8=\left(x-2\right)\left(x^3+3x^2-3x+4\right)\).

\(x^4-x^2-56\)

\(=x^4-8x^2+7x^2-56\)

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

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

\(x^4-x^2-56=x^4+7x^2-8x^2-56=x^2\left(x^2+7\right)-8\left(x^2+7\right)\)

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

\(x^4-x^2-56\)

\(=x^4+7x^2-8x^2-56\)

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

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

~ Rất vui vì giúp đc bn ~

     x^4 + x^2 + 1 

= x^4 + 2x^2   + 1 - x^2

= ( x^2  + 1)^2 - x^2

= ( x^2 - x + 1 )( x^2 + x + 1)