Đáp án: $ \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$
Giải thích các bước giải:
Ta có:
$\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$
$\to (\dfrac{a^2}{b+c}+a)+(\dfrac{b^2}{c+a}+b)+(\dfrac{c^2}{a+b}+c)=a+b+c$
$\to \dfrac{a^2+a(b+c)}{b+c}+\dfrac{b^2+b(c+a)}{c+a}+\dfrac{c^2+c(a+b)}{a+b}=a+b+c$
$\to \dfrac{a(a+b+c)}{b+c}+\dfrac{b(a+b+c)}{c+a}+\dfrac{c(a+b+c)}{a+b}=a+b+c$
$\to \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$ vì $a+b+c\ne 0$
