设x1,x2是方程x^2-4x+2=0的两根1/x1+/1x2= 绝对值x1-x2得

x1+x2=4
x1x2=2
所以:
(1/x1)+(1/x2)=(x1+x2)/(x1x2)=4/2=2
(x1-x2)^2=(x1+x2)^2-4x1x2=16-4*2=8
|x1-x2|=2*(根号2)
(x1+1)(x2+1)=x1x2+(x1+x2)+1=2+4+1=7