Question

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.

Answer 1

（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