Lời giải:
Nếu $a+b+c=0$ thì:
\(a+b=-c; b+c=-a; c+a=-b\)
\(\Rightarrow \left\{\begin{matrix} \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1(\text{thỏa mãn giả thiết})\\ P=\frac{-c}{2c}+\frac{-a}{3a}+\frac{-b}{4b}=\frac{-1}{2}+\frac{-1}{3}+\frac{-1}{4}=\frac{-13}{12}\end{matrix}\right.\)
Nếu $a+b+c\neq 0$. Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\)
\(\Rightarrow b+c=2a; c+a=2b; a+b=2c\)
\(\Rightarrow P=\frac{a+b}{2c}+\frac{b+c}{3a}+\frac{c+a}{4b}=\frac{2c}{2c}+\frac{2a}{3a}+\frac{2b}{4b}=1+\frac{2}{3}+\frac{1}{2}=\frac{13}{6}\)
