Question

cho a, b, c ∈ Z biết \(\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c}{c}\)

tính Q = \(\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)

Answer 1

\(\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c}{c}=\frac{a+b+c}{a+b+c}=1\)

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

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