∵x1是方程x²+x-1=0的根
∴x12+x1-1=0
即x12+x1=1
同理x22+x2=1
∴原式=(1-2)×(1-2)=1
化简整理得:
(x1^2+x1 -2)(x2^2 + x2 -2)=(x1x2)^2+x1x2(x1+x2) -2[(x1+x2)^2 -2x1x2] +4
有根与系数关系知:
x1+x2=-1 x1x2= -1代入上式即可计算出!
x1+x2=-b/a=-1,x1x2=c/a=-1
(x1²+x1-2)•(x2²+x2-2)
=(x1+2)(x1-1)(x2+2)(x2-1)
=【x1x2+2(x1+x2)+4】【x1x2-(x1+x2)+1】
=(-1-2+4)(-1+1+1)
=1×1
=1
将两根代入原方程即可得到要的式子,再代入待求式即可
(x1²+x1-2)•(x2²+x2-2)
=(x1+2)(x1-1)(x2+2)(x2-1)
=【x1x2+2(x1+x2)+4】【x1x2-(x1+x2)+1】
=(-1-2+4)(-1+1+1)
=1×1
=1