设 xa+yb=1, a+b=根号3, |ab|/根号a2+b2=1a2+b2=a2b2.a2+b2=(a+b)2-2ab=a2b2,代入 a+b=根号3,得(ab)2+2ab-3=0ab=1或-3.当ab=1时,由于 a+b=根号3,a>0,b>0.a+b=根号3≥根号2abab≤3/4.ab=1时无解,ab=-3,S= 1/2|ab|=3/2.