TH1: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\Rightarrow M=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\)
\(\Rightarrow M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
\(\Rightarrow M=\dfrac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}\)
\(\Rightarrow M=\dfrac{-abc}{abc}\)
\(\Rightarrow M=-1\)
TH2: a+b+c≠0
Áp dụng t/c dtsbn ta có;
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\Rightarrow M=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\)
\(\Rightarrow M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
\(\Rightarrow M=\dfrac{2a.2b.2c}{abc}\)
\(\Rightarrow M=\dfrac{8abc}{abc}\)
\(\Rightarrow M=8\)
