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Do a + b + c = 2016 suy ra: \(a=2016-\left(b+c\right);b=2016-\left(c+a\right);c=2016-\left(a+b\right)\)
Do đó:
\(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(c+a\right)}{c+a}+\frac{2016-\left(a+b\right)}{a+b}\)
\(=\frac{2016}{b+c}-1+\frac{2016}{c+a}-1+\frac{2016}{a+b}-1\)
\(=\left(\frac{2016}{b+c}+\frac{2016}{c+a}+\frac{2016}{a+b}\right)-3\)
\(=2016\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=2016.\frac{1}{6+2}-3=249\)
Vậy S = 249
N=(a/b+c)+(b/a+c)+(c/a+b)
N+3=(a/b+c)+1+(b/a+c)+1+(c/a+b)+1
N+3=(a+b+c/b+c)+(a+b+c/a+c)+(a+b+c/a+b)
N+3=(a+b+c)[(1/b+c)+(1/a+c)+(1/b+c)]
N+3=2016.(1/672)
N+3=3
=>N=0
\(N=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow N=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow N=\left(\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow N=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(\Rightarrow N=2016.\frac{1}{672}-3=0\)
Vậy N=0
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{c+a}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2016.\frac{1}{90}-3=\frac{97}{5}\)
Vậy....................
Lời giải:
\(\left\{\begin{matrix} a+b+c=2016\\ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{504}\end{matrix}\right.\)
\(\Rightarrow (a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2016.\frac{1}{504}=4\)
\(\Leftrightarrow \frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}=4\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}=4\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+1+1+1=4\)
\(\Leftrightarrow S+3=4\Leftrightarrow S=1\)
