Ta có:

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

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

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

\(\Rightarrow A>B\)

Vậy \(A>B\)

Ta có:$B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}$B=20002001+2002 +20012001−2002 

Vì:$\frac{2000}{2001}>\frac{2000}{2001+2002}$20002001 >20002001+2002 

$\frac{2001}{2002}>\frac{2001}{2001+2002}$20012002 >20012001+2002 

$\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)$⇒(20002001 +20012002 )>(20002001−2002 −20012001+2001 )

$\Rightarrow A>B$⇒A>B

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

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

nên \(\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\) hay A>B 

Ta có: 2000/2001>1/2 ;  2001/2002>1/2

=>A=1/2+1/2=1=>A>1

B=2000+2001/2001+2002=4001/4003<1

A>1;B<1

=>A>B

Vậy A>B

Theo mình nghĩ là :

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

Mà \(\frac{2000}{2001+2002}=\frac{2000}{4003}\)< \(\frac{2000}{2001}\)

\(\frac{2001}{2001+2002}\)= \(\frac{2001}{4003}\)<  \(\frac{2001}{2002}\)

Nên  : A > B

Không chắc nữa !!!

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

Ta có:

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

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

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

=> \(\left(\frac{2000}{2001}+\frac{2001}{2002}\right)\)> \(\left(\frac{2000}{2001+2002}+\frac{2001}{2001+2001}\right)\)

=> A>B

Vậy A>B

Ta có:B= \(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

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

Nên A>B

Ta có: 2000/2001>1/2 ;  2001/2002>1/2

=>A=1/2+1/2=1=>A>1

B=2000+2001/2001+2002=4001/4003<1

A>1;B<1

=>A>B

Vậy A>B

$B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}$B=20002001+2002 +20012001−2002 

Vì:

Ta có : \(B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}\)

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

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

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

\(\Rightarrow A>B\)

Ta có: B = \(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}=\frac{2000}{4003}+\frac{2001}{4003}\)

Ta thấy : \(\frac{2000}{2001}>\frac{2000}{4003}\)(1)

             \(\frac{2001}{2002}>\frac{2001}{4003}\) (2)

Từ (1) và (2) cộng vế với vế, ta được :

  \(\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{4003}+\frac{2001}{4003}\)

hay \(A=\frac{2000}{2001}+\frac{2001}{2002}>B=\frac{2000+2001}{2001+2002}\)

B=2000/2001+2002 + 2001/2001+2002

Ta có:2000/2001 > 2000/2001+2002

2001/2002 > 2001/2001+2002

Vậy A >B

I DON'T KNOW OKKKKK

2000 .... 2001