Question

初三一元二次方程

请问怎么解。答案是C。

Answer 1

m是方程的解
所以m²-2m-1=0
m²-2m=1
所以伏晌7m²-14m=7
同理
n²-2n=1
3n²-6n=3
所以尺厅清原式=(7+a)(3-7)=8
a=-9
所陵前以选C

Answer 2

x^2-2x-1=0,m,n是两个纤友桥告凯根
那么毁猛也满足m^2-2m-1=0 n^2-2n-1=0
m^2-2m=1 7m^2-14m=7
n^2-2n=1 3n^2-6n=3
原=(7+a)(3-7)=8
7+a=-2
a=-9
C

Answer 3

∵败纯大x^2-2x-1=0,m,n是两个裤并根察竖
∴有m^2-2m-1=0 n^2-2n-1=0
(7m^2-14m+a)(3n^2-6n-7)=8
(7m^2-14m-7+7+a)(3n^2-6n-3-4)=8
[7(m^2-2m-1)+7+a)][3(n^2-2n-1)-4]=8
(7×0+7+a)(3×0-4)=8
(7+a)(-4)=8
7+a=-2
a=-9
选C