Đáp án: $P=4$ hoặc $P=-4$
Giải thích các bước giải:
Nếu $a+b+c+d=0\to a+b=-(c+d)\to \dfrac{a+b}{c+d}=-1$
Tương tự $\to \dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}=-1$
$\to P=-4$
Nếu $a+b+c\ne 0$
Ta có:
$\dfrac{a+b+c-2020d}{d}=\dfrac{a+b+d-2020c}{c}=\dfrac{a+c+d-2020b}{b}=\dfrac{b+c+d-2020a}{a}$
$\to \dfrac{a+b+c+d-2021d}{d}=\dfrac{a+b+c+d-2021c}{c}=\dfrac{a+b+c+d-2021b}{b}=\dfrac{a+b+c+d-2021a}{a}$
$\to \dfrac{a+b+c+d}{d}-2021=\dfrac{a+b+c+d}{c}-2021=\dfrac{a+b+c+d}{b}-2021=\dfrac{a+b+c+d}{a}-2021$
$\to \dfrac{a+b+c+d}{d}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{a}$
$\to d=c=b=a$
$\to \dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}=1$
$\to P=4$
