Question

急用.11题,很急

Answer 1

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

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

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

=0/[(a-b)(a-c)(b-c)]

=0