* TH1: a + b + c + d \(\ne\)0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2d}{a+b+c+d}\)
\(=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow2a+b+c+d=5a;a+2b+c+d=5b\)
\(\Rightarrow b+c+d=3a;a+c+d=3b\)
\(\Rightarrow b+c+d+a+c+d=3a+3b\)
\(\Rightarrow\left(a+b\right)+2\left(c+d\right)=3\left(a+b\right)\)
\(\Rightarrow2\left(c+d\right)=2\left(a+b\right)\)
\(\Rightarrow c+d=a+b\)
CMTT ta được: \(b+c=a+d\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
* TH2: \(a+b+c+d=0\)
\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(d+a\right)\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)\(=-4\)
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