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

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

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

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

a, x^2 + 7x + 6

= x^2 + x + 6x + 6

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

= (x + 6)(x + 1)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

~ Rất vui vì giúp đc bn ~

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

\(=x^2y^2+xy-x^2y+xy^2+xy+1-x+y+x^2y+x-x^2+xy-xy^2-y+xy-y^2\)

\(=x^2y^2+2xy-x^2-y^2+1\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

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

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

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

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

x^2 + 7x -15

= x^2 + 7x +12,25  -27,25

= (x+3,5)^2 - 27, 25

= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))