Question

已知x1,x2是方程x^2-2x-2=0,的两个实数根，不解方程，求下列各式的值：

（1）1/x1+1/x2
(2)(x1+1)（x2+1)
谢了~

Answer 1

x1+x2=2
x1x2=-2
1/x1+1/x2=x1+x2/x1x2=2/-2=-1

(x1+1)(x2+1)=x1x2+x1+x2+1=-2+2+1=1

Answer 2

x1+x2=2
x1x2=-2
∴1/x1+1/x2
=（x1+x2)/（x1x2)
=2/(-2)
=-1
(2)(x1+1)（x2+1)
=x1+x2+x1x2+1
=2-2+1
=1