Question

如果m.n是两个不相等的实数,且满足m的平方减去2m等于1..n方减2n等于1.那么代数式2m方加4n方加1994等于多

Answer 1

m^2-2m=1

n^2-2n=1

m≠n∈R

可知m、n为方程x^2-2x=1的两个根

所以有：m+n=2

则m^2+n^2.(已知两个式子相加得）

=2m+2n+2

=2（m+n）+2

=6

所以把所求式子化简得

2m^2+4n^2-4n+1994

=2m^2+2n^2+2（n^2-2n）+1994

=2（m^2+n^2）+2（n^2-2n）+1994

=2*6+2*1+1994

=2008