Question

Cho a ; b ; c > 0 và : \(\dfrac{2b+c-a}{a}=\dfrac{2c+a-b}{b}=\dfrac{2a+b-c}{c}\) . Tính :

\(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{abc}\)

Answer 1

Vì \(a;b;c>0\) nên \(a+b+c>0\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{2b+c-a}{a}=\dfrac{2c+a-b}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c+a-b+2a+b-c}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}2b+c=3a\Leftrightarrow3a-2b=c\\2c+a=3b\Leftrightarrow3b-2c=a\\2a+b=3c\Leftrightarrow3c-2a=b\end{matrix}\right.\)

Khi đó: \(\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{abc}=\dfrac{abc}{abc}=1\)