C1: +Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}\)\(=\dfrac{2a+2b+2c}{a+b+c}\)
\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{b}{b}+\dfrac{a}{b}\right)\cdot\left(\dfrac{c}{c}+\dfrac{b}{c}\right)\cdot\left(\dfrac{a}{a}+\dfrac{c}{a}\right)\)
\(=\dfrac{b+a}{b}\cdot\dfrac{c+b}{c}\cdot\dfrac{a+c}{a}\)\(=\dfrac{b+a}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}\)\(=2\cdot2\cdot2=8\)
\(\Rightarrow M=8\)
C2:
+)Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b}{c}+1=\dfrac{b+c}{a}+1=\dfrac{c+a}{b}+1\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)
+)Vậy, ta có:
\(\dfrac{a+b+c}{c}=\dfrac{b+c+a}{c}\) và \(\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)
\(-\)Vì \(\dfrac{a+b+c}{c}=\dfrac{b+c+a}{c}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{b+c+a}{a}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\) (1)
\(-\)Vì \(\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)
\(\Rightarrow\dfrac{b+c+a}{a}-\dfrac{c+a+b}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\) (2)
\(-\)Mà theo đề bài ta có a,b,c đôi một khác 0 nên:
Từ (1) và(2) ta suy ra được:
\(\rightarrow\)a+b+c=0
\(\Rightarrow a+b=\left(-c\right)\)
\(\Rightarrow b+c=\left(-a\right)\)
\(\Rightarrow c+a=\left(-b\right)\)
+)Ta có:
M= \(\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
=\(\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{c}\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{c}=\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\) =(-1)
Vậy M= \(\left\{{}\begin{matrix}8\\-1\end{matrix}\right.\)
