\(\frac{2000\cdot2001-1001}{1999\cdot2002-999}=\frac{1999\cdot2001+2001-1001}{1999\cdot2001+1999-999}\)

\(=\frac{1999\cdot2001+1000}{1999\cdot2001+1000}=1\)

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

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:

\(\frac{2000}{2001}>\frac{2000}{2001+2002}\)

\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)

\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

\(\Rightarrow A>B\)

\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}

nguyen thieu cong thanh lm đúng rồi mà

=1/2000-1/2001+1/2001-1/2002+1/2002-1/2003+......+1/2009-1/2010

=1/2000-1/2010

=1/402000

\(\frac{1}{2000+2001}+\frac{1}{2001+2002}+\frac{1}{2002+2003}+...+\frac{1}{2009+2010}\)

\(=\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2003}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(=\frac{1}{2000}-\frac{1}{2010}\)

\(=\frac{1}{402000}\)

\(\frac{1}{2000}\)+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

\(\frac{1}{2000+2001}+\frac{1}{2001+2002}+\frac{1}{2002+2003}+...+\frac{1}{2009+2010}\)

\(=\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2002}-...+\frac{1}{2009}-\frac{1}{2010}\)

\(=\frac{1}{2000}-\frac{1}{2010}\)

\(=\frac{1}{402000}\)