Đơn giản biểu thức sau. 1/a(a-b)(a-c)+1/b(b-c)(b-a)+1/c(c-a)(c-b)

Ta có: 

1a(a−b)(a−c)+1b(b−c)(b−a)=b(b−c)−a(a−c)ab(a−b)(b−c)(a−c)=b2−bc−a2+acab(a−b)(b−c)(a−c)=(a−b)(−a−b+c)ab(a−b)(b−c)(a−c)=−a−b+cab(b−c)(a−c)(a≠b)

Vậy:

1a(a−b)(a−c)+1b(b−c)(b−a)+1c(c−a)(c−b)=−a−b+cab(b−c)(a−c)+1c(c−a)(c−b)=−ac−bc+c2+ababc(a−c)(b−c)=(a−c)(b−c)abc(a−c)(b−c)=1abc(a≠c,b≠c)