tính nhanh : 2001/2000-2002/2001
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{2000\cdot2001-1001}{1999\cdot2002-999}=\frac{1999\cdot2001+2001-1001}{1999\cdot2001+1999-999}\)
\(=\frac{1999\cdot2001+1000}{1999\cdot2001+1000}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
vì \(\frac{2000}{2001}>\frac{2000}{2001+2002};\frac{2001}{2002}>\frac{2001}{2001+2002}\)
=>A>B
Ta có: B=2000/2001+2002+2001/2001+2002
vì: 2000/2001>2000/2001+2002
2001/2002>2001/2001+2002
nên 2000/2001+2001/2002>2000/2001+2002+2001/2001+2002
Vậy A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{2000}{2001}+\frac{2001}{2002}\) và \(B=\frac{2000+2001}{2001+2002}\)
\(\Rightarrow B=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Ta Xét:
\(\frac{2000}{2001}>\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}
![](https://rs.olm.vn/images/avt/0.png?1311)
=1/2000-1/2001+1/2001-1/2002+1/2002-1/2003+......+1/2009-1/2010
=1/2000-1/2010
=1/402000
\(\frac{1}{2000+2001}+\frac{1}{2001+2002}+\frac{1}{2002+2003}+...+\frac{1}{2009+2010}\)
\(=\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2003}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=\frac{1}{2000}-\frac{1}{2010}\)
\(=\frac{1}{402000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2000}\)+2001+\(\frac{1}{2001}\)+ 2002+\(\frac{1}{2002}\)+2003+...+\(\frac{1}{2009}\)+2010
2001,0005+2002,0005+2003,0005+...+2010,0005
Số số hạng là:
(2010,0005-2001,0005)+1=10( số)
Số cặp số hạng là:
10:2= 5 ( cặp)
Tổng từng cặp là: 2001,0005+2010,0005=2002,0005+2009,0005=...=4011,001
Tổng của các số hạng trên là :
4011,001x5=20055,005
\(\frac{1}{2000+2001}+\frac{1}{2001+2002}+\frac{1}{2002+2003}+...+\frac{1}{2009+2010}\)
\(=\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2002}-...+\frac{1}{2009}-\frac{1}{2010}\)
\(=\frac{1}{2000}-\frac{1}{2010}\)
\(=\frac{1}{402000}\)
\(\frac{2001}{2000}-\frac{2002}{2001}=\frac{2001.2001}{2000.2001}-\frac{2002.2000}{2000.2001}\)
\(=\frac{\left(2002-1\right).\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.\left(2000+1\right)-\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+2002-2000-1-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+1-2002.2000}{2000.2001}\)
\(=\frac{1}{2000.2001}\)