Question

Упростить выражение - (a2/(a-b)(a-c))+(b2/(b-c)(b-a))+(c2/(c-a)(c-b))

Answer 1

$frac{ {a}^{2} }{(a - b)(a - c)} + frac{ {b}^{2} }{(b - c)( b - a)} + frac{ {c}^{2} }{(c - a)(c - b)} = frac{ {a}^{2} }{(a - b)(a - c)} - frac{ {b}^{2} }{(b - c)(a - b)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{ {a}^{2}(b - c) - {b}^{2} (a - c)}{(a - b)(a - c)(b - c)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{ {a}^{2}b - {a}^{2} c - {b}^{2} a + {b}^{2} c}{(a - b)(a - c)(b - c)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{ab(a - b) - c( {a}^{2} - {b}^{2} )}{(a - b)(a - c)(b - c)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{(a - b)(ab - c(a + b))}{(a - b)(a - c)(b - c)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{ab - ac - bc}{(a - c)(b - c)} + frac{ {c}^{2} }{(a - c)(b - c)} = frac{ab - ac - bc + {c}^{2} }{(a - c)(b - c)} = frac{b(a - c) - c(a - c)}{(a - c)( b - c)} = frac{(a - c)(b - c)}{(a - c)(b - c)} = 1$