(1)原式=x+y+z)(-x+y+z)(x-y+z)(x+y-z)
=[(x+y+z)(x+y-z)]*{[z+(x-y)][z-(x-y)]}
=[(x+y)^2-z^2][z^2-(x-y)^2]
=(x^2+2xy+y^2-z^2)[z^2-(x-y)^2]
=4*x^2*y^2-(x^2+y^2-z^2)^2
(2)原式
=x^2(x^2+1)+2x-1
=(3-√5)/2*[(3-√5)/2+1]+2*(√5-1)/2-1
=-1/2+√5-2
=(2√5-5)/2
(3)分解x^2+xy-2y^2=0
即：（x+2y)(x-y)=0
x=-2y或 x=y
(x^2+3xy+y^2)/(x^2+y^2)=[(x^2+xy-2y^2)+(2xy+3y^2)]/(x^2+y^2)
=(2xy+3y^2)/(x^2+y^2)
将x=-2y ,x=y分别代入可得
原式=-1/5或5/2