Ta có : \(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(\Rightarrow P+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{c+a}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow P+3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(\Rightarrow P+3=2019.10\)
\(\Rightarrow P+3=20190\)
\(\Rightarrow P=20190-3\)
\(\Rightarrow P=20187\)
Vậy P = 20187