已知|ab-2|与|b-1|互为相反数试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004
问题描述:
已知|ab-2|与|b-1|互为相反数试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004
它们的值,
答
∣ab-2∣与∣b-1∣互为相反数
那么∣ab-2∣+∣b-1∣=0
∣ab-2∣>=0与∣b-1∣>=0
那么∣ab-2∣=0与∣b-1∣=0
那么ab=2,b=1得到a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+ ……+1/(a+2004)(b+2004)
=1/1*2+1/2*3+.1/2005*2006
=1-1/2+1/2-1/3+.+1/2005-1/2006
=1-1/2006
=2005/2006