\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a-b+c}{b}+2=\dfrac{-a+b+c}{a}+2\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{a}\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(\circledast\) Với \(a+b+c=0\) thì \(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
\(\circledast\) Với \(a=b=c\) thì \(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{8a^3}{a^3}=8\)
Giải
Áp dung T/C.........
Ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)\(=\dfrac{a+b-c+a-b+c+\left(-a\right)+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\) a+b-c=c \(\Rightarrow\) a+b=2c
a-b+c=b a-c=2b
- a+b+c=a c+b=2a
\(\Rightarrow\) a+b-a-c=2c-2b
a-c-c+b=2b-2a
\(\Rightarrow\) b-c=2c-2b \(\Rightarrow\) 3b=3c
a-b=2b-2a 3a=3b
\(\Rightarrow\)a=b=c
Ta thay vào:
\(\dfrac{\left(a+b\right)\left(a+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)
