Question

Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)

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

( Gợi ý: Nhân cả 2 vế của đề bài với a + b + c )

Help me!!! @Phùng Khánh Linh

Answer 1

Có:

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

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

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

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

\(\Rightarrow\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c\)

\(\Rightarrow\)\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=a+b+c-a-b-c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\left(đpcm\right)\)