Đáp án: $M=\pm4$
Giải thích các bước giải:
Ta có:
$\dfrac{2019a+b+c+d}{a}=\dfrac{a+2019b+c+d}{b}=\dfrac{a+b+2019c+d}{c}=\dfrac{a+b+c+2019d}{d}$
$\to\dfrac{2018a+a+b+c+d}{a}=\dfrac{2018b+a+b+c+d}{b}=\dfrac{2018c+a+b+c+d}{c}=\dfrac{2018d+a+b+c+d}{d}$
$\to2018+\dfrac{a+b+c+d}{a}=2018+\dfrac{a+b+c+d}{b}=2018+\dfrac{a+b+c+d}{c}=2018+\dfrac{a+b+c+d}{d}$
$\to\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}$
$\to a+b+c+d=0$ hoặc $a=b=c=d$
Nếu $a+b+c+d=0$
$\to a+b=-(c+d)\to \dfrac{a+b}{c+d}=-1$
Tương tự $\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}=-1$
$\to M=-4$
Nếu $a=b=c=d\to M=4$
