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Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) (1)
Ta có : a+b+c khác 0
do nếu a+b+c=0=>\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\)=>-3=1(Vô lí)
do a+b+c khác 0 nên ta nhân (a+b+c) vào (1)
=>\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
=>\(\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c\left(a+b\right)+c^2}{a+b}=a+b+c\)
=>\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)(ĐPCM)
Có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\) (NHân cả hai vế vs a+b+c)
\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
=> đpcm
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{ac}{b+c}+\frac{bc}{c+a}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab+bc}{c+a}+\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+a+b+c=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
đpcm
ta có \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=\frac{\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=1\)