Question

一道初二的计算题,[（2a-b-c)/(a-b)(a-c)]+[(2b-c-a)/(b-c)(b-a)]+[(2c-a-

Answer 1

[（2a-b-c)/(a-b)(a-c)]+[(2b-c-a)/(b-c)(b-a)]+[(2c-a-b)/(c-b)(c-a）

=[(a-b)+(a-c)]/[(a-b)(a-c)]+[(b-c)+(b-a)]/[(b-c)(b-a)]+[(c-a)+(c-b)]/[(c-a)(c-b)]

=1/(a-c)+1/(a-b)+1/(b-a)+1/(b-c)+1/(c-b)+1/(c-a)

=1/(a-c)+1/(a-b)-1/(a-b)+1/(b-c)-1/(b-c)-1/(a-c)

=0