Question

cho a+b+c=2016 và \(\frac{1}{a+b}+\frac{1}{c+a}+\frac{1}{c+a}=\frac{1}{672}\) tính N=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Answer 1

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

Answer 2

\(N=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(\Rightarrow N=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(\Rightarrow N=\left(\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\right)-3\)

\(\Rightarrow N=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(\Rightarrow N=2016.\frac{1}{672}-3=0\)

Vậy N=0