x+y=3,xy=-2,试求代数式（1)(x-y）²; (2) x的四次方+y的四次方的值.

（1)(x-y）²
=x^2+y^2-2xy
=(x+y)^2-4xy
=3^2-4*(-2)
=9+8
=17
(2) x^4+y^4
=(x^2+y^2)^2-2x^2y^2
=[(x+y)^2-2xy]^2-2(xy)^2
=[3^2-2*(-2)]^2-2*(-2)^2
=(9+4)^2-8
=13^2-8
=169-8
=161