\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}\)
\(=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(=\dfrac{a^3+b^3+c^3-3abc+3abc}{abc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{3abc}\)
\(=\dfrac{3abc}{abc}=3\)
Vậy A = 3