Question

已知m.n是二元一次方程x^2-3x+1=0的两根,那么代数式2m^2+4n^2-6n+1999

Answer 1

m,n是方程的两根，那么
n^2-3n+1=0
m+n=3,mn=1
2m^2+4n^2-6n+1999
=2m^2+2n^2+2n^2-6n+1999
=2(m^2+n^2)+2(n^2-3n+1)+1997
=2[(m+n)^2-2mn]+1997
=2*(3^2-2)+1997
=14+1997
=2011

Answer 2

根据韦达定理，
m+n=3
mn=1

又n是方程的根，则n^2-3n+1=0 n^2-3n=-1

2m^2+4n^2-6n+1999
=2m^2+2n^2+2n^2-6n+1999
=2(m^2+n^2)+2(n^2-3n)+1999
=2[(m+n)^2-2mn]+2(-1)+1999
=2(9-2)-2+1999
=12+1999
=2011

Answer 3

根据韦达定理
m+n=3
mn=1
m²+n²=（m+n）²-2mn=7
n²-3n+1=0，即n²-3n=-1
所以
2m²+4n²-6n+1999
=2（m²+n²）+2（n²-3n）+1999
=2*7+2*（-1）+1999
=2011

Answer 4

2013
解法：
m+n=3
mn=1
m=1/n
n+1/n=3
n^2=3n-1 (1)
同样
m^2=3m-1 (2)
将（1）（2）带入2m^2+4n^2-6n+1999
得出6m-2+12n-4-6n+1999=6(m+n)-4=2013