Question

Cho \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\) Chứng minh rằng:\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)

Answer 1

Ta có: \(pt\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b\left(c+a\right)}{c+a}+\dfrac{c\left(a+b\right)}{a+b}=a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{bc}{c+a}+\dfrac{ba}{c+a}+\dfrac{ca}{a+b}+\dfrac{cb}{a+b}=a+b+c\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\) đpcm

Từ cái sau suy ra cái trước thì còn dễ hơn, đề này tui mà chưa làm kiểu kia sao bk làm :)