m,n是方程x2+3x+1=0的两根
由韦达定理得:m+n=-3
且
m²+3m+1=0
m²=-3m-1
同理:n²=-3n-1
则:2m2+4n2-6n+2000
=2(3m-1)+4(3n-1)-6n+2000
=6m-2+12n-4-6n+2000
=6m+6n+1994
=6(m+n)+1994
=6×(-3)+1994
=1994-18
=1976
m,n是方程x2+3x+1=0的两根
由韦达定理得:m+n=-3
且
m²+3m+1=0
m²=-3m-1
同理:n²=-3n-1
则:2m2+4n2-6n+2000
=2(3m-1)+4(3n-1)-6n+2000
=6m-2+12n-4-6n+2000
=6m+6n+1994
=6(m+n)+1994
=6×(-3)+1994
=1994-18
=1976