Đáp án: 1

Ta có: \({\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2}\)

\({a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right) = {a^2} + {b^2} + {c^2}\)

Suy ra \(2\left( {ab + bc + ac} \right) = 0\) hay \(ab + bc + ac = 0\).

Xét \({a^2} + 2bc = {a^2} + 2bc - ab - bc - ca = {a^2} - ab - ca + bc = \left( {a - b} \right)\left( {a - c} \right)\).

Tương tự ta có \({b^2} + 2ac = \left( {b - a} \right)\left( {b - c} \right);{\rm{ }}{c^2} + 2ab = \left( {c - a} \right)\left( {c - b} \right)\).

Do đó, ta có: \(P = \frac{{{a^2}}}{{{a^2} + 2bc}} + \frac{{{b^2}}}{{{b^2} + 2ac}} + \frac{{{c^2}}}{{{c^2} + 2ab}}\)

\( = \frac{{{a^2}}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \frac{{{b^2}}}{{\left( {b - a} \right)\left( {b - c} \right)}} + \frac{{{c^2}}}{{\left( {c - a} \right)\left( {c - b} \right)}}\)

\( = - \frac{{{a^2}\left( {b - c} \right) + {b^2}\left( {c - a} \right) + {c^2}\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}\)

\( = - \frac{{{a^2}b - {a^2}c + {b^2}c - {b^2}a + {c^2}a - {c^2}b}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}\)

\( = - \frac{{ab\left( {a - b} \right) - \left( {{a^2} - {b^2}} \right)c + {c^2}\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}\)

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

\( = - \frac{{\left( {a - b} \right)\left[ {a\left( {b - c} \right) - c\left( {b - c} \right)} \right]}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}\)

\( = - \frac{{\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}}\)

\( = \frac{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {c - a} \right)\left( {b - c} \right)}} = 1\).

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