Giải thích các bước giải:
Ta có :
$S+3=(\dfrac{a}{b+c}+1)+(\dfrac{b}{a+c}+1)+(\dfrac{c}{a+b}+1)$
$\to S+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}$
$\to S+3=(a+b+c)(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a})$
$\to S+3=2019.2019$
$\to S=2019^2-3$