1）已知分式3/（x^2+3x+1）=3/8,求1/（2x^2+6x-3）的值.2）已知x^2-5x-2009=0,求[（x-2）^3-(x-1)^2+1]/x-2的值.3）(2a-b-c)/[(a-b)(a-c)]+(2b-c-a)/[(b-c)(b-a)]+(2c-a-b)/[(c-a)(c-b)]

1) 1/11

1)2种方法.一种你可以先求X的值这样麻烦.但是保险
还有一种（x^2+3x+1）=8
(2x^2+6x+2)=16
(2x^2+6x+2-5)=11
2x^2+6x-3=11
SO 1/（2x^2+6x-3）=1/11
2）由已知x^2-5x-2009=0整理得：x^2-5x=2009,则有
[(x-2)^3-(x-1)^2+1]/(x-2)
=[(x-2)^3-(x^2-2x+1)+1]/(x-2)
=[(x-2)^3-x^2+2x]/(x-2)
=[(x-2)^3-x(x-2)]/(x-2)
=(x-2)×[(x-2)^2-x]/(x-2)
=(x-2)^2-x
=x^2-4x+4-x
=x^2-5x+4
=2009+4
=2013
3)原式=（2a-b-c)(b-c)/(a-b)(a-c)(b-c)-(2b-c-a)(a-c)/(a-b)(a-c)(b-c)+(2c-a-b)(a-b)/(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)
=2ab-b^2-bc-2ac+bc+c^2-2ab+ac+a^2+2bc-c^2-ac+2ac-a^2-2bc+2ac+ab+b^2/(a-b)(a-c)(b-c)
=2ac+ab/(a-b)(a-c)(b-c)
=a(2c+b)/(a-b)(a-c)(b-c)