Question

cho a, b, c khác 0 thoả mãn: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) . Tính S = \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)

Answer 1

Lời giải:

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

$\Leftrightarrow \frac{a^2}{b+c}+\frac{ab}{c+a}+\frac{ca}{a+b}+\frac{ab}{b+c}+\frac{b^2}{c+a}+\frac{bc}{a+b}+\frac{ac}{b+c}+\frac{bc}{c+a}+\frac{c^2}{a+b}=a+b+c$

$\Leftrightarrow S+\frac{ab+ac}{b+c}+\frac{ab+bc}{a+c}+\frac{bc+ac}{a+b}=a+b+c$

$\Leftrightarrow S+a+b+c=a+b+c$

$\Leftrightarrow S=0$