Cho a+b + c
Va 1/a+b +1/b+c + 1/ c+a = 1/90
Tinh S= a/b+c + b/c+a + c/a+b
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\(a,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0.abc=0\)
Mà \(a+b+c=1=>\left(a+b+c\right)^2=1=>a^2+b^2+c^2+2ab+2bc+2ac=1\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=1=>a^2+b^2+c^2=1-0=1\) (vì ab+bc+ac=0)
\(b,S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(=2014.\frac{1}{2014}-3=1-3=-2\)
Vậy.....................
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy....................
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2015.\frac{1}{90}-3=19\frac{7}{18}\)
\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(=2010.\frac{1}{3}=670\)
\(\Rightarrow S=670-3=667\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{1}{7}\left(a+b+c\right)\) (nhân a + b +c vào mỗi vế)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2009}{7}\)
Suy ra \(S=\frac{2009}{7}-3=284\)
a+b+c bằng mấy