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\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(=2010.\frac{1}{3}=670\)
\(\Rightarrow S=670-3=667\)
S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=>S+3=\(\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
=>S+3=\(\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
=>S+3=(a+b+c).\(\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
Thay a + b + c = 2011 và 1/(a+b) + 1/(b+c) + 1/(c+a) = 1/2010 vào S ta đc:
S+3=2011.1/2010
=>S=2011/2010-3
=>S=\(\frac{-4019}{2010}\)
Vậy S=-4019/2010 với a + b + c = 2011 và 1/(a+b) + 1/(b+c) + 1/(c+a) = 1/2010.
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy....................
a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b
=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1
= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1
= 2010.(1/b+c + 1/c+a + 1/a+b) - 3
= 2010.1/3 - 3 = 667
Vậy S = 667
Tk mk nha
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
=> (a+b+c)(1/a+b +a/b+c +1/c+a)=2010 . 1/2010
=>(a+b+c) /(a+b) +(a+b+c)/(b+c) + (a+b+c)/(a+c)=1
=> 1 +c/a+b + 1 +a/b+c + 1 +b/a+c=1
=>a/b+c +b/a+c +c/a+b =-2