设在[1,﹢∝)的任意两点x1和x2,x1＜x2f(x1)=x1+1/x1 f(x2)=x2+1/x2f(x1)-f(x2)=x1-x2+1/x1-1/x2=(x2-x1)[1/(x1·x2)-1]因为x1,x2≧1 则1/x1≤0 1/x2≤0 1/(x1·x2)-1＜0则f(x1)-f(x2)＜0又x1＜x2因此f(x)在该区间内...