1=(a+b+c)*(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=((a+b)+c)*(\frac{a+b}{ab}+\frac{1}{c})=\frac{(a+b)2 }{ab}+\frac{c*(a+b){ab}+\frac{a+b}{c}+1=(a+b)*(\frac{a+b}{ab}+\frac{c}{ab}+\frac{1}{c})+1=(a+b)*\frac{ac+bc+c2+ab}{abc}+1=\frac{(a+b)*(b+c)*(c+a)}{abc}+1\Rightarrow \frac{(a+b)*(b+c)*(c+a)}{abc}=0 \Rightarrow a=-b hoặc b=-c hoặc c=-a\Rightarrow A=0
$1=(a+b+c)*(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=((a+b)+c)*(\frac{a+b}{ab}+\frac{1}{c})
$$=\frac{(a+b)
^2 }{ab}+\frac{c*(a+b){ab
}}+\frac{a+b}{c}+1=(a+b)*(\frac{a+b}{ab}+\frac{c}{ab}+\frac{1}{c})+1
$$=(a+b)*\frac{ac+bc+c
^2+ab}{abc}+1=\frac{(a+b)*(b+c)*(c+a)}{abc}+1
$$\Rightarrow \frac{(a+b)*(b+c)*(c+a)}{abc}=0 \Rightarrow a=-b
$ hoặc
$b=-c
$ hoặc
$c=-a
$$\Rightarrow A=0
$