Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=-\dfrac{1}{c^3}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{-1}{c}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}-\dfrac{3}{abc}=\dfrac{-1}{c^3}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(\Rightarrow A=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)
\(=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
\(=\left(abc\right)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
=\(abc.\dfrac{3}{abc}\)
=3
Vậy A=3
