Vì  2008 2009 < 1;  10 9 > 1

Nên 2008 2009 <  10 9

a) \(y^2-x^2+6y+9\)

\(=\left(y^2+6y+9\right)-x^2\)

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

\(=\left[\left(y+3\right)-x\right]\left[\left(y+3\right)+x\right]\)

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

b) \(4y^2-x^2-4y+1\)

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

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

\(=\left[\left(2y-1\right)+x\right]\left[\left(2y-1\right)-x\right]\)

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

c)  \(\left(x-y\right)^2-x^2+y^2\)

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

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

\(=\left(x-y\right)\left[\left(x-y\right)-\left(x+y\right)\right]\)

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

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

d) \(x^6-y^6\)

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

\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)

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

a: =(y+3)^2-x^2

=(y+3+x)(y+3-x)

b: =(2y-1)^2-x^2

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

c: =x^2-2xy+y^2-x^2+y^2

=2y^2-2xy

=2y(y-x)

d: =(x^3-y^3)(x^3+y^3)

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

`a)x^2-2x+2+4y^2+4y`

`=x^2-2x+1+4y^2+4y+1`

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

`b)4x^2+y^2+12x+4y+13`

`=4x^2+12x+9+y^2+4y+4`

`=(2x+3)^2+(y+2)^2`

`c)x^2+17+4y^2+8x+4y`

`=x^2+8x+16+4y^2+4y+1`

`=(x+4)^2+(2y+1)^2`

`d)4x^2-12xy+y^2-4y+13`

`=4x^2-12x+9+y^2-4y+4`

`=(2x-3)^2+(y-2)^2`

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

b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)

c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)

d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)

Ta có x-y cùng tính chẵn lẻ với x-y

         y-z cùng tính chẵn lẻ với y-z

         z-x cùng tính chẵn lẻ với z-x

=>/x-y/+/y-z/+/z-x/ cùng tính chẵn lẻ với (x-y)+(y-z)+(z-x)=x-y+y-z+z-x=(x-x)+(y-y)+(z-z)=0, là 1 số chẵn

=>/x-y/+/y-z/+/z-x/ là 1 số chẵn

Vậy ko có x,y,z thỏa mãn đề bài