\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{b+c+d+a}{c+d+a}=\frac{c+d+a+b}{d+a+b}=\frac{d+a+b+c}{a+b+c}\)
Xét \(a+b+c+d\ne0\) suy ra \(a=b=c=d\). Khi đó:
\(P=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=1+1+1+1=4\)
Xét \(a+b+c+d=0\) suy ra \(\left\{\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(b+a\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\). Khi đó:
\(P=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
