Question

cho a,b,c là các số hữu tỉ khác 0 thỏa mãn:

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

Tính giá trị của biểu thức

A=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Answer 1

\(\dfrac{a+b-2c}{c}=\dfrac{c+a-2b}{b}=\dfrac{b+c-2a}{a}=\dfrac{a+b-2c+c+a-2b+b+c-2a}{c+b+a}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-2c}{c}=0\\\dfrac{c+a-2b}{b}=0\\\dfrac{b+c-2a}{a}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b-2c=0\\a+c-2b=0\\b+c-2a=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)