x2+(√3)y=√5,y2+(√3)x=√5,两式相减,
得(X-Y)*(X+Y)+√3(Y-X)=0,
即(X-Y)*(X+Y-√3)=0,
因为x≠y,所以X+Y=√3,所以X2+Y2+2XY=3,
又因为XY=1,所以X2+Y2=1,
(y/x)+(x/y)=X2+Y2/(XY)=1.
x2+(√3)y=√5,y2+(√3)x=√5,两式相减,
得(X-Y)*(X+Y)+√3(Y-X)=0,
即(X-Y)*(X+Y-√3)=0,
因为x≠y,所以X+Y=√3,所以X2+Y2+2XY=3,
又因为XY=1,所以X2+Y2=1,
(y/x)+(x/y)=X2+Y2/(XY)=1.