Question

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...

Answer 1

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

Answer 2

【(x²-y²)/2】²+xy
=[(x-y)(x+y)/2]²+xy
={[(1-√2+√3)/2-(1+√2-√3)/2][(1-√2+√3)/2+(1+√2-√3)/2]/2}²+(1-√2+√3)/2*(1+√2-√3)/2
={[(2√3-2√2)/2][(1+1)2]/2}²+[1-(√2-√3)²]/4
=(√3-√2)²+1/4-(2-2√6+3)/4
=3-2√6+2+1/4-5/4+√6/2
=4-3/2√6

Answer 3

（根号6-根号3+5根号2+1）/2