m,n是两个不相等的实数根,且满足m²-2m=1,n²-2n=1,那么代数式2m²+4n²-4n+1996=?
问题描述：
m,n是两个不相等的实数根,且满足m²-2m=1,n²-2n=1,那么代数式2m²+4n²-4n+1996=?
急,
答
由题意知m,n是方程x²-2x-1=0的两个不等根
∴m+n=2,mn=-1
2m²+4n²-4n+1996
=2m²+2n²+4mn-4mn+2n²-4n-2+2+1999
=2(m+n)²-4mn+2(n²-2n-1)+2001
=2*4+4+2001
=2013