设m,n是两个不相等的实数,且满足m^2-2m=1,n^2-2n=1,那么代数式2m^2+4n^2-4n+1991 的值为?如题,坐等.

问题描述:

设m,n是两个不相等的实数,且满足m^2-2m=1,n^2-2n=1,那么代数式2m^2+4n^2-4n+1991 的值为?
如题,坐等.

如题m n不相等 则m n为方程x^2-2x=1 方程的两个解 则有 m+n=2 mn=-1 设第一个方程为a1 第二个为a2 则 2(a1)+4(a2)=2m^2-4m+4n^2-8n=2m^2+4n^2-4(m+n)-4n=2m^2+4n^2-4n-8=6
则可得 2m^2+4n^2-4=14 2m^2+4n^2-4n+1991 =2005

m^2-2m=1,n^2-2n=1,m,n是两个不相等的实数;
m,n是方程x^2-2x-1=0,Δ>0;的两个根,维达定理,m+n=2,m*n=-1;
2m^2+4n^2-4n+1991 =2m^2+2n^2+2n^2-4n-2+1993
=2(m+n)^2-4mn+2(n^2-2n-1)+1993=8+4+1993=2005

2m^2+4n^2-4n+1991 最后的题目里没有m么?

m^2-2m=1,n^2-2n=1
所以m,n是方程x^2-2x=1的两个不相等的实数根。所以
m+n=2,mn=-1
2m^2+4n^2-4n+1991
=2(m^2+n^2)+2(n^2-2n)+1991
=2(m+n)^2-4mn+2*1+1991
=2005

那么m,n是方程x²-2x-1=0的两根
→m+n=2①,mn=-1②,n²-2n=1③,所以m²+n²=6④
所以2m²+4n²-4n=2(m²+n²+n²-2n)=2(6+1)=14
→结果为2005

123489

由m,n是两个不相等的实数,
且满足m^2-2m=1,
n^2-2n=1,
可知m,n是x^2-2x-1=0两个不相等的实数根。
则m+n=2,
又m^2=2m+1,
n^2=2n+1
2M^2+4N^2-4N+1994
=2(2M+1)+4(2N+1)-4N+1991
=4M+2+8N+4-4N+1991
=4(M+N)+1997
=4*2+1997
=2005