Question

2道初2分式计算题

1，已知b-1的相反数等于它本身，ab与-2互为相反数，求ab分之1+（a+1)(b+1)分之1+……+（a+2009)(b+2009)分之1的值 2，已知（x-1)(x-2)分之3x-4=x-1分之A+x-2分之B，求A,B的值 写出步骤，谢谢``

Answer 1

1，已知b-1的相反数等于它本身，ab与-2互为相反数，求ab分之1+（a+1)(b+1)分之1+……+（a+2009)(b+2009)分之1的值
-(b-1)=b-1,解得b=1
ab+(-2)=0,解得:a=2
原式=1/1*2+1/2*3+...+1/2010*2011
=1-1/2+1/2-1/3+...+1/2010-1/2011
=1-1/2011
=2010/2011
2，已知（x-1)(x-2)分之3x-4=x-1分之A+x-2分之B，求A,B的值
解:(3x-4)/((x-1)(x-2))=a/(x-1)+b/(x-2)
(3x-4)/((x-1)(x-2))=(a(x-2)+b(x-1))/((x-1)(x-2))
(3x-4)/((x-1)(x-2))=(ax-2a+bx-b)/((x-1)(x-2))
(3x-4)/((x-1)(x-2))=((ax+bx)-(2a+b))/((x-1)(x-2))
(3x-4)/((x-1)(x-2))=
((a+b)x-(2a+b))/((x-1)(x-2))
a+b=3
,2a+b=4
a=1
,b=2