Ta có: \(A=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
\(=\left(\dfrac{b+c}{a}+1\right)+\left(\dfrac{c+a}{b}+1\right)+\left(\dfrac{a+b}{c}+1\right)-3\)
\(=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-3\)
\(=\left(a+b+c\right).0-3=-3\)
Vậy A = -3
\(A=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
\(=\left(\dfrac{b+c}{a}+1\right)+\left(\dfrac{c+a}{b}+1\right)+\left(\dfrac{a+b}{c}+1\right)-3\)
\(=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-3\)
\(=\left(a+b+c\right).0-\dfrac{1}{3}=-\dfrac{1}{3}\)
Vậy \(A=-\dfrac{1}{3}\)
