(a+m)/(b+m)-(a/b)
=[b(a+m)-a(b+m)]/[b(b+m)]
=(ab+bm-ab-am)/[b(b+m)]
=(b-a)m/[b(b+m)]
已知a、b、m∈R+,且a<b
所以,b-a>0,b+m>0
所以,(b-a)m/[b(b+m)]>0
则,(a+m)/(b+m)-(a/b)>0
所以,(a+m)/(b+m)>a/
(a+m)/(b+m)-(a/b)
=[b(a+m)-a(b+m)]/[b(b+m)]
=(ab+bm-ab-am)/[b(b+m)]
=(b-a)m/[b(b+m)]
已知a、b、m∈R+,且a<b
所以,b-a>0,b+m>0
所以,(b-a)m/[b(b+m)]>0
则,(a+m)/(b+m)-(a/b)>0
所以,(a+m)/(b+m)>a/