题目

帮帮我忙： 9.化简1/(a+b)+1/(a-b)-(a-b)/(a^2+ab+b^2)-(a+b)/(a^2-ab+b^2)
14.已知a,b,c互不相等,求(2a-b-c)/[(a-b)(a-c)]+(2b-c-a)/[(b-c)(b-a)]+(2c-a-b)/[(c-a)(c-b)]

提问时间：2021-01-05

答案

9.化简1/(a+b)+1/(a-b)-(a-b)/(a^2+ab+b^2)-(a+b)/(a^2-ab+b^2)
=[1/(a+b)-(a+b)/(a^2-ab+b^2)]+[1/(a-b)-(a-b)/(a^2+ab+b^2)]
=[(a^2-ab+b^2)-(a+b)^2]/(a^3+b^3)+[(a^2+ab+b^2)-(a-b)^2]/(a^3-b^3)
=-3ab/(a^3+b^3)+3ab/(a^3-b^3)
=3ab[1/(a^3-b^3)-1/(a^3+b^3)]
=3ab[(a^3+b^3)-(a^3-b^3)]/(a^6-b^6)
=6ab^4/(a^6-b^6)
14.已知a,b,c互不相等,求(2a-b-c)/[(a-b)(a-c)]+(2b-c-a)/[(b-c)(b-a)]+(2c-a-b)/[(c-a)(c-b)]
=(2a-b-c)/[(a-b)(a-c)]-(2b-c-a)/[(b-c)(a-b)]+(2c-a-b)/[(a-c)(b-c)]
=[(2a-b-c)(b-c)-(2b-c-a)(a-c)+(2c-a-b)(a-b)]/[(a-b)(a-c)(b-c)]
=0/[(a-b)(a-c)(b-c)]
=0