\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}-2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{2}=\dfrac{b+c-a}{a}+\dfrac{2a}{a}=\dfrac{c+a-b}{b}+\dfrac{2b}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)
\(\Rightarrow a=b=c\) Thay vào P ta được:
\(P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8.a^3}{a^3}=8\)
Vậy, \(P=8\).
ta có \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Rightarrow\dfrac{a+b}{c}-1=\dfrac{b+c}{a}-1=\dfrac{c+a}{b}-1\)
\(\Rightarrow\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
+) với a + b + c = 0
\(\Rightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
+) với a + b + c ≠ 0
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)
