Question

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

chứng minh rằng \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

 

Answer 1

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

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

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

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

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

= (a+b+c) - (a+b+c) = 0 

Answer 2

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