Giải thích các bước giải:
Ta có :
$\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1$
$\to \dfrac{a+b}{c}-1=\dfrac{b+c}{a}-1=\dfrac{c+a}{b}-1=1$
$\to \dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=2$
$\to \dfrac{a+b}{c}.\dfrac{b+c}{a}.\dfrac{c+a}{b}=2.2.2=8$
$\to \dfrac{a+b}{a}.\dfrac{b+c}{b}.\dfrac{c+a}{c}=8$
$\to (1+\dfrac ba)(1+\dfrac cb)(1+\dfrac ac)=8$
$\to M=8$