TH1:
\(\dfrac{a}{b+c+d}=\dfrac{b}{c+d+a}=\dfrac{c}{d+a+b}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=a+c+d\\3c=a+b+d\\3d=a+b+c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3\left(a-b\right)=b-a\\3\left(b-c\right)=c-b\\3\left(c-d\right)=d-c\\3\left(d-a\right)=a-d\end{matrix}\right.\) \(\Rightarrow a=b=c=d\)
\(\Rightarrow P=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}=1+1+1+1=4\)
TH2: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a-b}{a-a}=-1\)
\(\Rightarrow-a=b+c+d\Rightarrow a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\a+c=-\left(b+d\right)\\a+d=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{b+c}{-\left(b+c\right)}+\dfrac{c+d}{-\left(c+d\right)}+\dfrac{-\left(b+c\right)}{b+c}=-1+-1+-1+-1=-4\)
Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)
