Dạng 2: Đơn giản biểu thức.

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)

Do a+b+c=0=>a+b=−c=>a2+2ab+b2=c2

Nên a2+b2−c2=−2abb2+c2−a2=−2bca2+c2−b2=−2ac

Vậy 1a2+b2−c2+1b2+c2−a2+1c2+a2−b2

=−12ab+12bc+12ca=−a+b+c2abc