Question

初二分式的加减法2a-b-c/（a-b)(a-c)+ 2b-c-a/(b-c)(b-a)+ 2c-b-a/(c-b)(c

Answer 1

1、

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

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

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

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

=0

2、

原式=(3a-6b-4a+5b)/(a+b)-(5a-6b-7a+8b)/(a-b)

=(-a-b)/(a+b)-(-2a+2b)/(a-b)

=-(a+b)/(a+b)+2(a-b)/(a-b)

=-1+2

=1