Ta có :
$\dfrac{3}{a+b} = \dfrac{2}{b+c} = \dfrac{1}{c+a}$
$\Rightarrow \dfrac{a+b}{3} = \dfrac{b+c}{2}= \dfrac{c+a}{1}$
Đặt $\dfrac{a+b}{3} = \dfrac{b+c}{2}= \dfrac{c+a}{1} = k$
Suy ra :
$a+ b = 3k$
$b+c = 2k$
$a+c = k$
Suy ra :
$(a+b) - (b+c) = 3k - 2k ⇒ a - c = 1k$
⇒ $a = (k+k) : 2 = k$
⇒ $c = 0$
⇒ $b = 2k$
Khi đó :
$P = \dfrac{3a+3b+2019c}{a+b- 2020c} = \dfrac{9k}{3k} = 3$
Vậy $P = 3$
