Question

求代数式x的四次方-y的四次方/x³+x²y+xy²+y³的值,期中x=2002

Answer 1

原式=(x²+y²)(x²-y²)/[x²(x+y)+y²(x+y)]

=(x²+y²)(x+y)(x-y)/[(x+y)(x²+y²)]

=x-y

=2002-2001

=1