Question

Cho \(a+b+c=0.\)Tính \(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)\)

Answer 1

Ta có: \(a+b+c=0\)

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

\(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)\)

\(=\left(1+\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{a}{c}\right).\left(1+\dfrac{c}{a}\right)\)

\(=\left(1+\dfrac{a+b}{c}+\dfrac{a}{b}\right).\left(1+\dfrac{c}{a}\right)\)

\(=1+\dfrac{c}{a}+\dfrac{a+b}{c}+\dfrac{a+b}{a}+\dfrac{a}{b}+\dfrac{c}{b}\)

\(=1+\dfrac{a+b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)

= 1 + 0 -1 -1 (vì a+b+c=0; a+c=-b; a+b=-c)

= -1