Question

Cho a+b+c=14 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\) . Tính \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Answer 1

ta có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)

\(\Rightarrow14.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=14.\frac{1}{7}\)

\(\Rightarrow\frac{14}{a+b}+\frac{14}{b+c}+\frac{14}{c+a}=2\)

mà a+b+c =14

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=2\)

\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{a+c}{a+c}+\frac{b}{a+c}\right)=2\)

\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2\)

\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2-3\)

\(\Rightarrow A=-1\)

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