> 数学 >
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
人气:157 ℃ 时间:2019-12-03 00:33:07
解答
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版