Question

若x=二分之一减（根号二加根号三）,y=二分之（一加根号二减根号三）,求

Answer 1

应该是x=[1-(√2+√3)]/2,y=(1+√2-√3)/2吧

因1=(√3)^2-(√2)^2=(√3-√2)(√3+√2)

则x=[(√3-√2)(√3+√2)-(√2+√3)]/2=[(√3+√2)(√3-√2-1)]/2=-(√3+√2)y

y=[(√3-√2)(√3+√2)+√2-√3)]/2=[(√3-√2)(√3+√2-1)]/2=-(√3-√2)x（或由上式变形）

于是x^2/y^2+y^2/x^2=[-(√3+√2)]^2+[-(√3-√2)]^2=10,即x^4+y^4=10x^2y^2

从而[(x^2-y^2)/2]^2=x^4+y^4-2x^2y^2=8x^2y^2

又x=[(1-√3)-√2]/2

y=[(1-√3)+√2]/2

则xy={[(1-√3)-√2]*[(1-√3)+√2]}/4=[(1-√3)^2-(√2)^2]/4=(1-√3)/2

x^2y^2=[(1-√3)/2]^2=(2-√3)/2

所以[(x^2-y^2)/2]^2=8*(2-√3)/2=8-4√3