Question

 

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

            CM : \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\) 

Answer 1

Ta có :

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

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

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

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

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

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a+b+c-a-b-c=0\)

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