如果m、n是两个不相等的实数,且满足m2-2m=1,n2-2n=1,那么代数式2m2+4n2-4n+1994=_.

问题描述:

如果m、n是两个不相等的实数,且满足m2-2m=1,n2-2n=1,那么代数式2m2+4n2-4n+1994=______.

根据题意可知m,n是x2-2x-1=0两个不相等的实数根.
则m+n=2,
又m2-2m=1,n2-2n=1
2m2+4n2-4n+1994
=2(2m+1)+4(2n+1)-4n+1994
=4m+2+8n+4-4n+1994
=4(m+n)+2000
=4×2+2000
=2008.