Question

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

Answer 1

(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