Question

cho a+b+c=2017 và \(\frac{1}{a+b}\) + \(\frac{1}{b+c}\)+ \(\frac{1}{c+a}\)= \(\frac{1}{10}\). 

Tính M= \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)

Answer 1

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)

\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017.\frac{1}{10}\)

\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)

Mà \(2017=a+b+c\) nên :

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\)

\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{b}{a+c}\right)=201,7\)

\(3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7\)

\(\Leftrightarrow M=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7-3\)

\(\Leftrightarrow M=198,7\)

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Answer 2

198,7 do ban a