cho a,b,c,d là các số thực thỏa mãn a+b+c-2020d/d=a+b+d-2020c/c=a+c+d-2020b/b=b+c+d-2020a/a Tính P=a+b/c+d +b+c/d+a +c+d/a+b +d+a/b+c

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1. 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:

   $\dfrac{a+b+c-2020d}{d}=\dfrac{a+b+d-2020c}{c}=\dfrac{a+c+d-2020b}{b}=\dfrac{b+c+d-2020a}{a}$

   $\to \dfrac{a+b+c+d-2021d}{d}=\dfrac{a+b+c+d-2021c}{c}=\dfrac{a+b+c+d-2021b}{b}=\dfrac{a+b+c+d-2021a}{a}$

   $\to \dfrac{a+b+c+d}{d}-2021=\dfrac{a+b+c+d}{c}-2021=\dfrac{a+b+c+d}{b}-2021=\dfrac{a+b+c+d}{a}-2021$

   $\to \dfrac{a+b+c+d}{d}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{a}$

   $\to d=c=b=a$

   $\to \dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}=1$

   $\to P=4$

   Trả lời