Question

[Nguyễn Tiến Đạt]

cho a+b+c=2007 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\)=\(\frac{1}{90}\)

tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Answer 1

[ST]

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{90}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2017}{90}\)

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2017}{90}\)

\(\Rightarrow A+3=\frac{2017}{90}\)

\(\Rightarrow S=\frac{2017}{90}-3=\frac{1747}{90}\)

Answer 2

[vũ tiền châu]

từ giả thiết, ta có 

\(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}=\frac{1}{90}\)

Mà \(S=\frac{a}{2017-a}+\frac{b}{2017-b}+\frac{c}{2017-c}=-3+\frac{2017}{2017-a}+\frac{2017}{2017-b}+\frac{2017}{2017-c}\)

=-3+\(2017\left(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}\right)=-3+\frac{2017}{90}=\frac{1747}{90}\)

vậy ...

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