Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Rightarrow\frac{1}{a}-\frac{1}{a+b+c}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{a+b+c-a}{a\left(a+b+c\right)}+\frac{c+b}{bc}=0\)\(\Leftrightarrow\frac{b+c}{a\left(a+b+c\right)}+\frac{b+c}{bc}=0\)
\(\Leftrightarrow\left(b+c\right)\left(\frac{1}{a\left(a+b+c\right)}+\frac{1}{bc}\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(\frac{bc+a\left(a+b+c\right)}{abc\left(a+b+c\right)}\right)=0\)\(\Rightarrow\left(b+c\right)\left(bc+a^2+ab+ac\right)=0\)
\(\Rightarrow\left(b+c\right)\left[b\left(c+a\right)+a\left(c+a\right)\right]=0\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
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