Tính \(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\left(1\right)\)

Ta có :

\(\dfrac{b+c-3a}{a}=\dfrac{a+c-3b}{b}=\dfrac{a+b-3c}{c}\)

\(\Rightarrow\dfrac{b+c-3a}{a}+3=\dfrac{a+c-3b}{b}+3=\dfrac{a+b-3c}{c}+3\)

\(\Rightarrow\dfrac{b+c-3a}{a}+\dfrac{3a}{a}=\dfrac{a+c-3b}{b}+\dfrac{3b}{b}=\dfrac{a+b-3c}{c}+\dfrac{3c}{c}\)

\(\Rightarrow\dfrac{b+c-3a+3a}{a}=\dfrac{a+c-3b+3b}{b}=\dfrac{a+b-3c+3c}{c}\)

\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta được :

\(\Rightarrow\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b}{c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)Suy ra :

\(\left\{{}\begin{matrix}\dfrac{b+c}{a}=2\\\dfrac{a+c}{b}=2\\\dfrac{a+b}{c}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b+c}{2}=a\left(2\right)\\\dfrac{a+c}{2}=b\left(3\right)\\\dfrac{a+b}{2}=c\left(4\right)\end{matrix}\right.\)

THAY (2),(3),(4)vào(1) ta được :

M=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{b+c}{2}.\dfrac{a+c}{2}.\dfrac{a+b}{2}}\)

\(=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{2.2.2}}=2.2.2=2^3=8\)

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