Lời giải

Ta có:

\(P = \frac{{ab}}{{{c^2}}} + \frac{{bc}}{{{a^2}}} + \frac{{ac}}{{{b^2}}}\)

\(P = \frac{{abc}}{{{c^3}}} + \frac{{abc}}{{{a^3}}} + \frac{{abc}}{{{b^3}}} = abc\left( {\frac{1}{{{c^3}}} + \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}}} \right)\)

Vì \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\)

\( \Leftrightarrow \frac{1}{a} + \frac{1}{b} = \frac{{ - 1}}{c}\)

\( \Leftrightarrow {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} = {\left( {\frac{{ - 1}}{c}} \right)^3}\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{3}{{ab}}\left( {\frac{1}{a} + \frac{1}{b}} \right) = \frac{{ - 1}}{{{c^3}}}\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} + \frac{3}{{ab}}\left( {\frac{{ - 1}}{c}} \right) = 0\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} - \frac{3}{{abc}} = 0\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}}\) (1)

Thay (1) vào P ta được: \(P = abc.\frac{3}{{abc}} = 3\).