由不等式ax^2+bx-2>0的解为x1/3,可得x1=-1/2和x2=1/3是方程ax^2+bx-2=0的两实根.则ax^2+bx-2=a(x+1/2)(x-1/3)=ax^2+1/6ax-1/6a,方程两边对应系数相等,则可得方程组：b=1/6a,2=1/6a.解得：a=12,b=2.于是我们有ab=24....