Đáp án:
\[\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\]
\[\to (a+b+c)\bigg(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\bigg) = a+b+c\]
\[\to \dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c\]
\[\to \dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\]
\[\to P=0\]