Theo đề, ta có:
\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}=\dfrac{a+b-c+b+c-a+c+a-b}{3c+3a+3b}=\dfrac{a+b+c}{3c+3a+3b}\)
\(=\dfrac{a+b+c}{3.\left(a+b+c\right)}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{a+b-c}{3c}=\dfrac{1}{3}\Rightarrow a+b-c=\dfrac{3c}{3}=c\Rightarrow a+b=2c\)
và \(\dfrac{b+c-a}{3a}=\dfrac{1}{3}\Rightarrow b+c-a=\dfrac{3a}{3}=a\Rightarrow b+c=2a\)
và \(\dfrac{c+a-b}{3b}=\dfrac{1}{3}\Rightarrow c+a-b=\dfrac{3b}{3}=b\Rightarrow c+a=2b\)
\(\Rightarrow P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)
\(=\left(\dfrac{2c}{a}\right)\left(\dfrac{2b}{c}\right)\left(\dfrac{2c}{b}\right)=\dfrac{2c.2a.2b}{a.b.c}=8\)
Vậy P = 8
