Giải thích các bước giải:
Ta có :
$\dfrac{\overline{ab}}{a+b}=\dfrac{10a+b}{a+b}=\dfrac{10(a+b)-9b}{a+b}=10-\dfrac{9b}{a+b}$
Vì $0\le b\le 9,1\le a\le 9\to \dfrac{9b}{a+b}\le \dfrac{9b}{b+1}=\dfrac{9(b+1)-9}{b+1}=9-\dfrac{9}{b+1}\le 9-\dfrac{9}{9+1}=\dfrac{81}{10}$
$\to \dfrac{\overline{ab}}{a+b}\ge 10-\dfrac{81}{10}=\dfrac{19}{10}$
Dấu = xảy ra khi $a=1,b=9$