假设m＞n,m、n∈Rf(m)-f(n)={a-[2/(2^m+1)]}-{a-[2/(2^n+1)]}=-2[1/(2^m+1)-1/(2^n+1)]=-2{(2^n-2^m)/[(2^m+1)(2^n+1)]}=-2{2^n*[1-2^(m-n)]/[(2^m+1)(2^n+1)]}=负{正*负/正}＞0所以f(x)在R上是增函数...