Question

((a^2-bc)/((a-b)(a-c)))+((b^2+ac)/((b+c)(b-a)))+((c^2+ab)/((c-a)(c+b))) упростить

Answer 1

$\frac{ {a}^{2} - bc }{(a - b)(a - c)} + \frac{ {b}^{2} + ac}{(b + c)(b - a)} + \frac{ {c}^{2} + ab }{(c - a)(c + b)} = \frac{ {a}^{2} - bc }{(a - b)(a - c)} - \frac{ {b}^{2} + ac}{(b + c)(a - b)} - \frac{ {c}^{2} + ab }{(a - c)(b + c)} = \frac{( {a}^{2} - bc)(b + c) - ( {b}^{2} + ac)(a - c) }{(a - b)(a - c)(b + c)} - \frac{ {c}^{2} + ab}{(a - c)(b + c)} = \frac{ {a}^{2} b - {b}^{2} c + {a}^{2}c - b {c}^{2} - {b}^{2} a - {a}^{2}c + {b}^{2} c + a {c}^{2} }{(a - b)(a - c)(b + c)} - \frac{ {c}^{2} + ab}{(a - c)(b + c)} = \frac{ {a}^{2} b - b{c}^{2} - {b}^{2}a + a {c}^{2} }{(a - b)(a - c)(b + c)} - \frac{ {c}^{2} + ab}{(a - c)(b + c)} = \frac{ab(a - b) + {c}^{2}(a - b) }{(a - b)(a - c)(b + c)} - \frac{ {c}^{2} + ab}{(a - c)(b + c)} = \frac{(a - b)(ab + {c}^{2}) }{(a - b)(a - c)(b + c)} - \frac{ {c}^{2} + ab }{(a - c)(b + c)} = \frac{ab + {c}^{2} }{(a - c)(b + c)} - \frac{ {c}^{2} + ab}{(a - c)(b + c)} = 0$