Th1: \(a+b+c+d\ne0\)
\(\dfrac{2019a+b+c+d}{a}=\dfrac{a+2019b+c+d}{b}=\dfrac{a+b+2019c+d}{c}=\dfrac{a+b+c+2019d}{d}\)
\(\Rightarrow\dfrac{2019a+b+c+d}{a}-2018=\dfrac{a+2019b+c+d}{b}-2018=\dfrac{a+b+2019c+d}{c}-2018=\dfrac{a+b+c+2019d}{d}-2018\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}=1+1+1+1=4\)
Th2: \(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}=-1-1-1-1=-4\)
