来源:无涯教育2024-06-24 09:17:100
解:

(2a-b-c)/(a^2-ab-ac+bc)=[(a-b)+(a-c)]/(a-b)(a-c)=1/(a-b)+1/(a-c)
(2b-c-a)/(b^2-bc-ab+ac)=[(b-a)+(b-c)]/(b-a)(b-c)=1/(b-a)+1/(b-c)
(2c-a-b)/(c^2-ac-bc+ab)=[(c-a)+(c-b)]/(c-a)(c-b)=1/(c-a)+1/(c-b)

原式=1/(a-b)+1/(a-c)+1/(b-a)+1/(b-c)+1/(c-a)+1/(c-b)
=1/(a-b)+1/(b-a)+1/(a-c)+1/(c-a)+1/(b-c)+1/(c-b)
=0
2.原式=(b-c)^2/(a-b)(a-c)(b-c)+(c-a)^2/(b-a)(c-a)(b-c)+(a-b)^2/(c-a)(c-b)(a-b)

=(b-c)^2/(a-b)(a-c)(b-c)+(c-a)^2/(a-b)(a-c)(b-c)+(a-b)^2/(a-c)(b-c)(a-b)
=2(a^2+b^2+c^2-ab-ac-bc)/(a-b)(a-c)(b-c)
3.1/x-1/y=(y-x)/xy=3
∴x-y=-3xy
原式=(3x+5xy-3y)/(x-3xy-y)=(-9xy+5xy)/(-3xy-3xy)=2/3
4.(x-2)^3-(x-1)^2+1=(x-2)^3-x(x-2)=(x-2)(x^2-5x+4)
∵x^2-5x-2003=0
∴x^2-5x+4=2007
原式=(x-2)*2007/(x-2)=2007
5.原式=a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)/a
∵a+b+c=0
∴a+c=-b,a+b=-c,b+c=-a
∴(a+c)/b+(a+b)/c+(b+c)/a=-1+-1+-1=-3