Question

已知m,n是方程x^2-2x-1=0的两根，且(3m^2-6m+a)(5n^2-10n-8)=6,求a的值

Answer 1

(3m^2-6m+a)(5n^2-10n-8)=6
3（m²-2m+1/3a)×5（n²-2n-1.6）=6
由题意知道：m²-2m=n²-2n=1，带入上式
15×（1+1/3*a)×（1-1.6）=6
化简-9-3a=6，得a=-5

Answer 2

(3m^2-6m+a)(5n^2-10n-8)
=(3(m²-2m-1)+a+3)(5(n²-2n-1)-3)
=(0+a+3)(0-3)
=-3(a+3)=6
所以a+3=-2
即a=-5

Answer 3

解：因为m,n是方程x^2-2x-1=0的两根
所以m=1-2^(1/2)
n=1+2^(1/2)
代入3m^2-6m+a)(5n^2-10n-8)=6
a=-5

Answer 4

m,n是方程x^2-2x-1=0的两根，
m^2-2m-1=0,
m^2-2m＝1；
n^2-2n-1=0,
n^2-2n＝1；

故有(3m^2-6m+a)(5n^2-10n-8)
＝（3（m^2-2m）+a）(5(n^2-2n)-8)
=6;
(3+a)(5-8)=6;
a=-5;