a=1/(1/1000+1/1001+1/1002+.+1/1998+1/1999).求a的整数部分是多少
2个回答
a>1/(1/1000+1/1000+...+1/1000)(1000个)=1;
a
相关问题
(2000-1)+(1999-2)+(1998-3)+…+(1002-999)+(1001-1000)
(2000-1)+(1999-2)+(1998-3)+…+(1002-999)+(1001-1000)
| 1∕1001-1∕1000|+|1/1002-1/1001|=?
1/1001减去1/1000的绝对加上1/1002减去1/1001的绝对值减去1/1002减去1/1001的绝对值是多少
瞧一瞧看一看~设a=1/(1/1000+1/1001+1/1003+1/1004+……+1/1998+1/1999),则
求证1-1/2+1/3-1/4.+1/1999-1/2000=1/1001+1/1002+.+1/2000
(1/2000-1)*(1/1999-1)*…*(1/1001-1)*(1/1000-1)
求(1/2001-1)*(1/2000-1)*(1/1999-1)*…(1/1001-1)*(1/1000-1)的值
(1998/1-1)(1997/1-1)(1996/1-1)...(1001/1-1)(1000/1-1)
(1998/1-1)(1997/1-1)(1996/1-1).(1001/1-1)(1000/1-1).