Tính A=(1+a/b).(1+b/c).(1+c/a) biết (2016c-a-b)/c=(2016b-a-c)/b=(2016a-b-c)/a

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{2016c-a-b}{c}=\frac{2016b-a-c}{b}=\frac{2016a-b-c}{a}=\frac{2016c-a-b+2016b-a-c+2016a-b-c}{a+b+c}=\frac{2016\left(a+b+c\right)-2\left(a+b+c\right)}{a+b+c}=\frac{2014\left(a+b+c\right)}{a+b+c}=2014\)

\(\Rightarrow\left\{\begin{matrix}\frac{2016c-a-b}{c}=2014\\\frac{2016b-a-c}{b}=2014\\\frac{2016a-b-c}{a}=2014\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}2016c-a-b=2014c\\2016b-a-c=2014b\\2016a-b-c=2014a\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}-a-b=2014c-2016c\\-a-c=2014b-2016b\\-b-c=2014a-2016a\end{matrix}\right.\)

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

\(\Rightarrow\left\{\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\) (1)

Ta có \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(\Leftrightarrow A=\frac{a+b}{b}.\frac{c+b}{c}.\frac{a+c}{a}\)

Thế (1) vào biểu thức ta có :

\(A=\frac{a+b}{b}.\frac{c+b}{c}.\frac{a+c}{a}\)

\(\Rightarrow A=\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}\)

\(\Rightarrow A=2.2.2=8\)

Vậy biểu thức A=8