∵x1、x2是一元二次方程2x^2-3x-1=0的两个根
∴x1+x2=3/2,x1x2=-1/2
∴x2/x1+x1/x2
=（x2²+x1²）/x1x2
=[(x1+x2)²-2x1x2]/x1x2
=[(3/2)²-2*(-1/2)]/(-1/2)
=-13/2
(x1-2)(x2-2)
=x1x2-2(x1+x2)+4
=-1/2-2*(3/2)+4
=1/2
(x1+1/x2)(x2+1/x1)
=x1x2+1+1+1/(x1x2)
=-1/2+2+1/(-1/2)
=-1/2