若|ab-2|+|b-1|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+·若|ab-2|+|b-1|=0求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+······+1/(a+2004)(b+2004)的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 10:54:50
若|ab-2|+|b-1|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+·若|ab-2|+|b-1|=0求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+······+1/(a+2004)(b+2004)的值
x){ѽ&1^mXcklc=I DS6H6r4mj) T, jyڰ&H Nm Xڎ/R!dkdkFzO?o Phge32Z@ (n ҭ `c>(P.kقl<;P()9C

若|ab-2|+|b-1|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+·若|ab-2|+|b-1|=0求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+······+1/(a+2004)(b+2004)的值
若|ab-2|+|b-1|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+·
若|ab-2|+|b-1|=0
求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+······+1/(a+2004)(b+2004)的值

若|ab-2|+|b-1|=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+·若|ab-2|+|b-1|=0求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+······+1/(a+2004)(b+2004)的值
ab=2,b=1所以a=2.原式=1/(1*2) 1/(2*3) 1/(3*4) ... 1/(2005*2006)=(1-1/2) (1/2-1/3) ... (1/2005-1/2006)=1-1/2006=2005/2006