Question

跪求数学题：已知m.n是方程x2-2x-1=0的两根,求代数式(7m2-14m-3)(3n2-6n+500)的值

Answer 1

因为m、n是方程的两个根，所以有m+n=2，mn=-1
原代数式=21（mn）^2-42mn^2-9n^2-42nm^2+84mn+18n+3500m^2-14*500m-1500
=21-42mn(m+n)-9n^2+3500m^2+84mn-14*500m+18n-1500
=21+84-84+3500m^2-9n^2-7000m+18n-1500
=21+3500m(m-2)-9n(n-2)-1500
=21+3500m(-n)-9n(-m)-1500
=21+3500-9-1500

Answer 2

m.nê?·?3ìx2-2x-1=0μ?á??ù?ùò? m^2 - 2m - 1 = 0, m^2 - 2m = 1 í?àí n^2 - 2n = 1 ?ùò? (7m2-14m+a)(3n2-6n-7) = [7(m

Answer 3

代数式（7m2-14m-3)(3n2-6n+500）=（7m2-14m-7+4)(3n 2-6n-3+503）因为m,n是方程的解所以7m2-14m-7和3n2-6n-3都等于零，所以代数式等于4乘503=2012

Answer 4

2012