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\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a + b + c + d khác 0) nên a = b = c = d
\(\Rightarrow\frac{2a-b}{c+d}+\frac{2b-c}{d+a}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{1}{2}.4=2\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\)
\(\Rightarrow a=b=c=d\)
Khí đó:
\(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=4\)
Vậy M = 4
Vì \(\frac{a}{b+c+d}\)= \(\frac{b}{a+c+d}\)= \(\frac{c}{a+b+d}\)= \(\frac{d}{a+b+c}\)nên
\(\frac{a}{b+c+d}\)+1 = \(\frac{b}{a+c+d}\)+1 = \(\frac{c}{a+b+d}\)+1 = \(\frac{d}{a+b+c}\) +1
hay\(\frac{a+b+c+d}{b+c+d}\) = \(\frac{a+b+c+d}{a+c+d}\)= \(\frac{a+b+c+d}{a+b+d}\)= \(\frac{a+b+c+d}{a+b+c}\)
Mà a + b + c + d \(\ne\)0 \(\Rightarrow\) \(b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) \(M=4\)
\(\Leftrightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\frac{a}{a+b+c+d}=\frac{b}{a+b+c+d}=\frac{c}{a+b+c+d}=\frac{d}{a+b+c+d}\)
\(\Rightarrow a=b=c=d\) Thay vào A ta được :
\(A=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
Giải: Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a + b + c + d\(\ne\)0)
=> \(\frac{a}{b}=1\)=> a = b
\(\frac{b}{c}=1\) => b = c
\(\frac{c}{d}=1\) => c = d
\(\frac{d}{a}=1\) => d = a
=> a = b = c = d
Khi đó, ta có: \(\frac{2a-b}{c+d}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}\)
hay \(\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}+\frac{2a-a}{a+a}\)
\(=\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}+\frac{a}{2a}\)
= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
= \(\frac{1}{2}.4=2\)
Ta có:\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{a+b+d}{c}+1=\frac{a+b+c}{d}+1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Vì a+b+c+d\(\ne\)0=>a=b=c=d
\(\Rightarrow A=\frac{a+c}{b+d}+\frac{a+b}{c+d}+\frac{a+c}{b+d}+\frac{b+c}{a+d}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)