\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=\dfrac{a+b}{a}\cdot\dfrac{b+c}{b}\cdot\dfrac{c+a}{a}\\ P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài toán trở thành: - Hoc24
Th1: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
\(\Rightarrow P=\left(\dfrac{a+b}{a}\right)\left(\dfrac{b+c}{b}\right)\left(\dfrac{a+c}{c}\right)\)
\(\Rightarrow P=\dfrac{-c}{a}.\dfrac{-a}{b}.\dfrac{-b}{c}\)
\(\Rightarrow P=\left(-1\right).\left(-1\right).\left(-1\right)\)
\(\Rightarrow P=-1\)
Th1: a+b+c≠0
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow a+c=2b\)
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
\(\Rightarrow P=\dfrac{2c}{a}.\dfrac{2a}{b}.\dfrac{2b}{c}\)
\(\Rightarrow P=2.2.2\)
\(\Rightarrow P=8\)
