若x=(1-根号2+根号3)/2,y=(1+根号2-根号3)/2,求【(x²-y²)/2】²+xy的值...

若x=(1-根号2+根号3)/2,y=(1+根号2-根号3)/2
x+y=1
x-y=根号3-根号2
xy=[1-(根号2-根号3)]/2*[1+(根号2-根号3)]/2=[1-(根号2-根号3)^2]/4=(2根号6-4)/4=根号6/2-1
【(x²-y²)/2】²+xy
=(根号3-根号2)^2/4+根号6/2-1
=5/4-根号6/2+根号6/2-1
=1/4