Question

Cho dãy tỉ số bằng nhau

\(\frac{2018a+b+c+d}{a}\) = \(\frac{a+2018b+c+d}{b}\)= \(\frac{a+b+2018c+d}{c}\)= \(\frac{a+b+c+2018d}{d}\)

Tính M = \(\frac{a+b}{c+d}\)+ \(\frac{b+c}{d+a}\)+ \(\frac{c+d}{a+b}\)+ \(\frac{d+a}{b+c}\)

Answer 1

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Answer 2

dux trung doan ĐÃ GIÚP THÌ TỬ TẾ VÀO

Answer 3

ta xét 2 trường hợp 

th1:a+b+c+d khác 0

\(\frac{2018a+b+c+d}{a}=\frac{a+2018b+c+d}{b}\)

\(=\frac{2018c+a+b+d}{c}=\frac{2018d+a+b+c}{d}\)

\(2017+\frac{a+b+c+d}{a}=2017+\frac{a+b+c+d}{b}\)

\(=2017+\frac{a+b+c+d}{c}=2017+\frac{a+b+c+d}{d}\)

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

\(\Rightarrow a=b=c=d\)

thay vào bt M ta có

\(M=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{b+c}=1.4=4\)

Th2: a+b+c+d=0

=> a+b=-(c+d)

b+c=-(a+d)

thay vào bt m ta có
\(M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}=\left(-1\right).4=-4\)