Vì a + b + c = 2018
\(\Rightarrow\left\{{}\begin{matrix}b+c=2018-a\\c+a=2018-b\\a+b=2018-c\end{matrix}\right.\)
Ta có: \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a}{2018-a}+\dfrac{b}{2018-b}+\dfrac{c}{2018-c}\)
\(P+3=\left(\dfrac{a}{2018-a}+1\right)+\left(\dfrac{b}{2018-b}+1\right)+\left(\dfrac{c}{2018-c}+1\right)=\dfrac{2018}{b+c}+\dfrac{2018}{c+a}+\dfrac{2018}{a+b}=2018\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+c}\right)=2018.\dfrac{2017}{2018}=2017\Rightarrow P=2014\)
Ta có : \(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\)
\(\Rightarrow3+P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\)
\(\Rightarrow3+P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a +b+c}{a+b}\)
\(\Rightarrow3+P=\left(a+b+c\right).\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\)
Mà \(a+b+c=2018;\) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\) \(\left(a,b\in R\right)\)
\(\Rightarrow3+P=2018.\dfrac{2017}{2018}\)
\(\Rightarrow3+P=2017\)
\(\Rightarrow P=2014\)
Vậy \(P=2014\)
Cách khác nè :))
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{2018}\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=2017\)
\(\Leftrightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}+3=2017\)
\(\Leftrightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=2014\)
