Question

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)]

Answer 1

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

Answer 2

1/(a+b)+1/(a-b)-(a-b)/(a^2+ab+b^2)-(a+b)/(a^2-ab+b^2)
=1/(a+b)+1/(a-b)-(a-b)²/(a-b)(a²+ab+b²)-(a+b)²/(a+b)(a²-ab+b²)
=1/(a+b)+1/(a-b)-(a-b)²/(a-b)³-(a+b)²/(a+b)³
=1/(a+b)+1/(a-b)-1/(a-b)-1/(a+b)
=0
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)]
原式=(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-a-c)(a-c)+(2c-a-b)(a-b)]/(a-b)(a-c)(b-c)
=(2ab-b²-bc-2ac+bc+c²-2ab+a²+ac+2bc-ac-c²+2ac-a²-ab-2bc+ab+b²)/(a-b)(a-c)(b-c)
=0/(a-b)(a-c)(b-c)
=0

Answer 3

(2a-b-c)/[(a-b)(a-c)]=[(a-b)+(a-c)]/[(a-b)(a-c)=1/(a-c) +1/(a-b)
同理(2b-c-a)/[(b-c)(b-a)]=1/(b-a)+1/(b-c) ; (2c-a-b)/[(c-a)(c-b)]=1/(c-b)+1/(c-a)
但1/(a-c)与1/(c-a); 1/(a-b)与1/(b-a) ; 1/(b-c) ;1/(c-b)是相反数
=>和=0...ans

Answer 4

(2a-b-c)/[(a-b)(a-c)]=[(a-b)+(a-c)]/[(a-b)(a-c)=1/(a-c) +1/(a-b)
同理(2b-c-a)/[(b-c)(b-a)]=1/(b-a)+1/(b-c) ; (2c-a-b)/[(c-a)(c-b)]=1/(c-b)+1/(c-a)
但1/(a-c)与1/(c-a); 1/(a-b)与1/(b-a) ; 1/(b-c) ;1/(c-b)是相反数
=>和=0...ans