\[\begin{array}{l}
\frac{{2019a + b + c + d}}{a} = \frac{{a + 2019b + c + d}}{b} = \frac{{a + b + 2019c + d}}{c} = \frac{{a + b + c + 2019d}}{d} = \frac{{2022(a + b + c + d)}}{{a + b + c + d}} = 2022\\
 \to 2019a + b + c + d = 2022a;a + 2019b + c + d = 2022b;a + b + 2019c + d = 2022c;a + b + c + 2019d = 2022d\\
 \to a = b = c = d\\
 \to \frac{{a + b}}{{c + d}} + \frac{{b + c}}{{d + a}} + \frac{{c + d}}{{a + b}} + \frac{{d + a}}{{b + c}} = 1 + 1 + 1 + 1 = 4
\end{array}\]

Đáp án+Giải thích các bước giải:

`(2019a+b+c+d)/a=(a+2019b+c+d)/b=(a+b+2019c+d)/c=(a+b+c+2019d)/d`$\\$`=>(2019a+b+c+d)/a-2019=(a+2019b+c+d)/b-2019=(a+b+2019c+d)/c-2019=(a+b+c+2019d)/d-2019`$\\$`=>(a+b+c+d)/a=(a+b+c+d)/b=(a+b+c+d)/c=(a+b+c+d)/d(***)`

`**Nếu\ a+b+c+d=0=>a+b=-(c+d)=>(a+b)/(c+d)=-1`$\\$`**Nếu\ a+b+c+d=0=>b+c=-(a+d)=>(b+c)/(a+d)=-1`$\\$`**Nếu\ a+b+c+d=0=>c+d=-(a+b)=>(c+d)/(a+b)=-1`$\\$`**Nếu\ a+b+c+d=0=>d+a=-(b+c)=>(d+a)/(b+c)=-1`$\\$`=>M=-4`

`**Nếu \ a+b+c+d\ne0`$\\$`Từ\ (***)=>a=b=c=d=>(a+b)/(c+d)=(b+c)/(d+a)=(c+d)/(a+b)=(d+a)/(b+c)=1`$\\$`=>M=4`