Ta có: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{a^2b^2+b^2c^2+c^2a^2}{\left(abc\right)^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{a^2b^2+b^2c^2+c^2a^2}{64}=\frac{3}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=48\)
Vậy: \(A=\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}=\frac{48}{8}=6\)