\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{\left(3a+b+c\right)+\left(a+3b+c\right)+\left(a+b+3c\right)}{a+b+c}\)
\(=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\frac{3a+b+c}{a}=5\Rightarrow3a+b+c=5a\Rightarrow b+c=2a\)
Tương tự ta có : \(a+c=2b;a+b=2c\)
\(\Rightarrow B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}\)
\(=\frac{8abc}{abc}=8\)
Ta có:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{3c+b+a}{c}\)
\(\Rightarrow3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{b+a}{c}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}\)
TH1:\(a+b+c=0\)\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=-1\)
TH2:\(a+b+c\ne0\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\b+a=2c\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
Vậy \(\left[{}\begin{matrix}B=-1\\B=8\end{matrix}\right.\)