Ta có:
\(\frac{3}{a+b}=\frac{2}{b+c}=\frac{1}{c+a}.\)
\(\Rightarrow\frac{a+b}{3}=\frac{b+c}{2}=\frac{c+a}{1}.\)
Đặt \(\frac{a+b}{3}=\frac{b+c}{2}=\frac{c+a}{1}=k\Rightarrow\left\{{}\begin{matrix}a+b=3k\\b+c=2k\\c+a=1k\end{matrix}\right.\)
Có \(a+b+b+c+c+a=3k+2k+1k\)
\(\Rightarrow2a+2b+2c=\left(3+2+1\right).k\)
\(\Rightarrow2.\left(a+b+c\right)=6k\)
\(\Rightarrow a+b+c=6k:2\)
\(\Rightarrow a+b+c=3k.\)
\(\Rightarrow c=3k-a-b\)
\(\Rightarrow c=3k-3b\)
\(\Rightarrow c=0.\)
Lại có: \(P=\frac{3a+3b+2019c}{a+b-2020c}\)
\(\Rightarrow P=\frac{3a+3b+2019.0}{a+b-2020.0}\)
\(\Rightarrow P=\frac{3a+3b+0}{a+b-0}\)
\(\Rightarrow P=\frac{3a+3b}{a+b}\)
\(\Rightarrow P=\frac{3.\left(a+b\right)}{a+b}\)
\(\Rightarrow P=3.\)
Vậy \(P=3.\)
Chúc bạn học tốt!
