BhregvftevuyfvgfwygfyfwRrwWUwwefferewgfgugrgidmmcjfkdmekeodkkwmek=1029848483388882888388
ta xét 2 trường hợp
th1:a+b+c+d khác 0
\(\frac{2018a+b+c+d}{a}=\frac{a+2018b+c+d}{b}\)
\(=\frac{2018c+a+b+d}{c}=\frac{2018d+a+b+c}{d}\)
\(2017+\frac{a+b+c+d}{a}=2017+\frac{a+b+c+d}{b}\)
\(=2017+\frac{a+b+c+d}{c}=2017+\frac{a+b+c+d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
thay vào bt M ta có
\(M=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{b+c}=1.4=4\)
Th2: a+b+c+d=0
=> a+b=-(c+d)
b+c=-(a+d)
thay vào bt m ta có
\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}=\left(-1\right).4=-4\)