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\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{90}\Leftrightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{a+b+c}{a+b}\)\(\Leftrightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1=\dfrac{2007}{90}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{193}{10}\)
\(\Rightarrow S=\dfrac{193}{10}\)
\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{2016-\left(b+c\right)}{b+c}+\dfrac{2016-\left(c+a\right)}{c+a}+\dfrac{2016-\left(a+b\right)}{a+b}\)\(\Rightarrow S=2016\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(\Rightarrow S=2016.\dfrac{1}{90}-3\)
\(\Rightarrow S=\dfrac{97}{2}\)
Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)
\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)
Vậy A = 2013
Nhân cả hai vế của đẳng thức cho a+b+c ta được
\(\dfrac{a+b+c}{a+b}\)+\(\dfrac{a+b+c}{a+b}\)=\(\dfrac{a+b+c}{c+a}\)=\(\dfrac{a+b+c}{90}\)
=> a+ \(\dfrac{c}{a+b}\)+1+\(\dfrac{a}{b+c}\)+1+\(\dfrac{b}{c+a}\)=\(\dfrac{2007}{90}\)
=>\(\dfrac{a}{b+c}\)+\(\dfrac{b}{c+a}\)+\(\dfrac{c}{a+b}\)=\(\dfrac{2007}{90}\)-3= 22,3-3=19,3