已知Iab+2I+Ia+1I=0,求代数式1/(a-1)*(b+1)+1/(a-2)*(b+2)+…+1/(a-2004)*(b+2004)的值

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已知Iab+2I+Ia+1I=0,求代数式1/(a-1)*(b+1)+1/(a-2)*(b+2)+…+1/(a-2004)*(b+2004)的值

由|ab+2|+|a+1|=0得:ab=-2 ,a=-1; b=2 (a-1)(b+1)=-2x3,(a-2)(b+2)=-3x4,.(a-2004)(b+2004)=-2005x2006 原式=-(1/2-1/3+1/3-1/4+.+1/2005-1/2006)=-(1/2-1/2006)=1/2006-1/2=-1002/2006=-501/1003