Giải:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
\(\Leftrightarrow-\dfrac{a+b}{ab}-\dfrac{1}{c}+\dfrac{1}{a+b+c}=0\)
\(\Leftrightarrow-\dfrac{a+b}{ab}-\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Vậy ...
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
⇔ bc(a+b+c) + ac(a+b+c) + ab(a+b+c) = abc (quy đồng và khử mẫu vì a,b,c ≠ 0)
\(\Leftrightarrow abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc=abc\)
\(\Leftrightarrow bc\left(b+c\right)+a\left(c^2+2bc+b^2\right)+a^2\left(b+c\right)=0\)(chuyển abc ở vế phải sang chỉ còn 2abc rồi đặt nhân tử chung)
\(\Leftrightarrow\left(b+c\right)\left(bc+ab+ac+a^2\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left[b\left(a+c\right)+a\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+c\right)\left(a+b\right)=0\left(đpcm\right)\)
