Question

cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\) chứng minh rằng \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)

Answer 1

Lời giải:

Ta có:

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

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

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

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

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

Ta có đpcm.