Question

.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=?(过程)

Answer 1

M，N是方程x²-2x-1=0的两个根。
x²-2x=1
x²-2x+1=2
(x-1)²=2
x-1=±√2
x=1±√2
2M2+4N2-4N+1994
=4N2-4N+1+2M2+1993
=(2N-1)²+2M2+1993
当N=1+√2，M=1-√2时
=(2+2√2-1)²+2(1-√2)²+1993
=(1+2√2)²+2(1-2√2+2)+1993
=(1+4√2+8)+2(3-2√2)+1993
=9+4√2+6-4√2+1993
=15+1993
=2008
当N=1-√2，M=1+√2时
=(2-2√2-1)²+2(1+√2)²+1993
=(1-2√2)²+2(1+2√2+2)+1993
=(1-4√2+8)+2(3+2√2)+1993
=9-4√2+6+4√2+1993
=15+1993
=2008

Answer 2

你好
1.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=?(过程)
由m，n是两个不相等的实数，
且满足m2－2m＝1，
n2－2n＝1，
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