1X2/1=1--ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1.+(a+2005)(b+2005)/1的值如何求?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/04 01:30:54
1X2/1=1--ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1.+(a+2005)(b+2005)/1的值如何求?

1X2/1=1--ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1.+(a+2005)(b+2005)/1的值如何求?
1X2/1=1--
ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1.+(a+2005)(b+2005)/1的值如何求?

1X2/1=1--ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1.+(a+2005)(b+2005)/1的值如何求?
已知|ab-2|与|b-1|互为相反数,试求 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2003)(b+2003) 因为|ab-2]+|b-1|=0 所以:ab-2=0,b-1=0 b=1 a=2 原式=1/1*2+1/2*3+1/3*4+.+1/2004*2005 =1-1/2+1/2-1/3+1/3-1/4+.+1/2004-1/2005 =1-1/2005 =2004/2005

0