Cho a, b, c đôi một khác nhau và khác 0 thỏa mãn (a + b + c)² = a² + b² + c². Tính giá trị biểu thức \(A = \frac{{{a^2}}}{{{a^2} + 2bc}} + \frac{{{b^2}}}{{{b^2} + 2ca}} + \frac{{{c^2}}}{{{c^2} + 2ab}}\).

Ta có (a + b + c)² = a² + b² + c².

⇔ a² + b² + c² + 2(ab + bc + ca) = a² + b² + c².

⇔ 2(ab + bc + ca) = 0.

⇔ ab + bc + ca = 0.

⇔ bc = –ab – ca.

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

Chứng minh tương tự, ta được \(\frac{{{b^2}}}{{{b^2} + 2ca}} = \frac{{{b^2}}}{{\left( {b - a} \right)\left( {b - c} \right)}}\); \(\frac{{{c^2}}}{{{c^2} + 2ab}} = \frac{{{c^2}}}{{\left( {a - c} \right)\left( {b - c} \right)}}\).

Khi đó ta có \(A = \frac{{{a^2}}}{{{a^2} + 2bc}} + \frac{{{b^2}}}{{{b^2} + 2ca}} + \frac{{{c^2}}}{{{c^2} + 2ab}}\).

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

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

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

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

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

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

Vậy A = 1.