Cho a, b, c đôi một khác nhau thỏa mãn (a + b + c)² = a² + b² + c².

Tính $P = \frac{{{a^2}}}{{{a^2} + 2bc}} + \frac{{{b^2}}}{{{b^2} + 2ac}} + \frac{{{c^2}}}{{{c^2} + 2ab}}$.

Lời giải

Ta có (a + b + c)² = a² + b² + c²

⇔ a² + b² + c² + 2ab + 2bc + 2ac = a² + b² + c²

⇔ 2ab + 2bc + 2ac = 0

⇔ ab + bc + ac = 0

⇔ \(\left\{ \begin{array}{l}{\rm{ab =   - bc - ac}}\\{\rm{bc =   - ab - ac}}\\{\rm{ac =   - ab - bc}}\end{array} \right.\)

Thay \(\left\{ \begin{array}{l}{\rm{ab =   - bc - ac}}\\{\rm{bc =   - ab - ac}}\\{\rm{ac =   - ab - bc}}\end{array} \right.\) vào biểu thức P ta có

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

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

P = $\frac{{{a^2}}}{{{a^2} + bc - ab - ac}} + \frac{{{b^2}}}{{{b^2} + ac - ab - bc}} + \frac{{{c^2}}}{{{c^2} + ab - bc - ac}}$

P = $\frac{{{a^2}}}{{a(a - b) - c(a - b)}} + \frac{{{b^2}}}{{ - b(a - b) + c(a - b)}} + \frac{{{c^2}}}{{ - c(b - c) + a(b - c)}}$

P = $\frac{{{a^2}}}{{(a - b)(a - c)}} + \frac{{{b^2}}}{{(a - b)(c - b)}} + \frac{{{c^2}}}{{(b - c)(a - c)}}$

P = $\frac{{{a^2}(c - b) + {b^2}(a - c) + {c^2}(b - a)}}{{(a - b)(b - c)(c - a)}}$

P = $\frac{{{a^2}c - {a^2}b + {b^2}a - {b^2}c + {c^2}b - {c^2}a}}{{(a - b)(b - c)(c - a)}}$

P = $\frac{{{a^2}c - {a^2}b + {b^2}a - {b^2}c + {c^2}b - {c^2}a}}{{(a - b)(b - c)(c - a)}}$

P = $\frac{{(abc - {b^2}c - a{c^2} + b{c^2}) + ({a^2}c - {a^2}b + a{b^2} - abc)}}{{(a - b)(b - c)(c - a)}}$

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

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

P = $\frac{{(a - b)(b - c)(c - a)}}{{(a - b)(b - c)(c - a)}} = 1$

Vậy P = 1.