Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2019a+b+c+d}a+\frac{a+2019b+c+d}b+\frac{a+b+2019c+d}c+\frac{a+b+c+2019d}d=\frac{2022a+2022b+2022c+2022d}{a+b+c+d}=2022\)
Xét \(\frac{2019a+b+c+d}a=2022\)\(\Leftrightarrow 2019a+b+c+d=2022a
\Leftrightarrow 3a=b+c+d\) (1)
Xét \(\frac{a+2019b+c+d}b=2022\)\(\Leftrightarrow a+2019b+c+d=2022b
\Leftrightarrow 3b=a+c+d\) (2)
Xét \(\frac{a+b+2019c+d}c=2022\)\(\Leftrightarrow a+b+2019c+d=2022a
\Leftrightarrow 3c=a+b+d\) (3)
Xét \(\frac{a+b+c+2019d}d=2022\)\(\Leftrightarrow a+b+c+2019d=2022d
\Leftrightarrow 3d=a+b+c\) (4)
Từ (1)(2)(3)(4)\(\Rightarrow a=b=c=d\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)
\(M=1+1+1+1=4\)
Chúc bạn học tốt!
