设A(x1,3/x1)B(x2,5/x2)
AD= 3/x1= 5/x2,可得 x1= 3/5*x2
CD= x2-x1= 2/5*x2
矩形ABCD的面积
= AD*CD= 5/ x2 * 2/5*x2 = 2