Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=-3\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=9\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=9\)
Mà \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=7\)nên \(2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Rightarrow\frac{c}{abc}+\frac{b}{abc}+\frac{a}{abc}=1\)\(\Rightarrow\frac{1}{A}=\frac{a+b+c}{abc}=1\Rightarrow A=1\)