Question

[Yui Arayaki]

Cho a, b, c là ba số thực khác 0 thỏa mãn:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) và \(a+b+c\ne0\)

Hãy tính giá trị của biểu thức \(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\)

Answer 1

[Mashiro Shiina]

Lời giải \(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

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

\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)

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

Khi \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\Leftrightarrow B=\dfrac{-abc}{abc}=-1\)

Khi \(a=b=c\Leftrightarrow B=\dfrac{8abc}{abc}=8\)