1、若m,n满足m+n=-2005,mn=7,求(m²+2004m+6)(n²+2006n+8)的值

1、若m,n满足m+n=-2005,mn=7,求(m²+2004m+6)(n²+2006n+8)的值.
2、设x(x-1)-(x²-y)=-2,求1/2(x²+y²)-xy的值.

(m²+2004m+6)(n²+2006n+8)
=[m(m+2004)+6][n(n+2006)+8]
=[m(-n-1)+6][n(-m+1)+8]
=(-mn+6-m)(-mn+n+8)
=-(1+m)(n+1)
=-(mn+m+n+1)
=-(7-2005+1)
=1997
x(x-1)－(x²－y)=－2
x^2-x-x^2+y=-2
x-y=2
(x²＋y²)/2－xy
=1/2(x²＋y²-2xy)
=(x-y)^2/2
=2^2/2
=4/2
=2