Sai thì bỏ qua ( bạn bè mà ) !
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1-1-1=-3\)(vô lí )
\(\Rightarrow a+b+c\ne0\)
Ta có :
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)
Đặt a + b + c = H
\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{a+c}+\frac{ac}{a+b}+\frac{b^2}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{c^2}{b+a}+\frac{ac}{c+b}+\frac{bc}{a+c}=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\left(\frac{ac}{a+b}+\frac{bc}{a+b}\right)+\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+a+b+c=H\)( Chỗ này làm hơi tắt bỏ qua nha )
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=H-\left(a+b+c\right)\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=0\left(đpcm\right)\)
ĐK:....
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)(nhân vào rồi tách)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Việt Hoàng _ TTH (*Yonko Team*): Mình chưa xem kỹ nhưng có lẽ hướng làm bạn là sai òi nhé!
