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

A>B vì tử số của B cộng với 2001 và mẫu số cộng thêm 2002 

k cho tui với nhé

tíck cho tui với

A = \(\frac{2000+2001}{2001+2002}\)= \(\frac{4001}{4003}\)

B = \(\frac{2000+2001}{2001+2003}=\frac{4001}{4003}\)

vậy A = B

A=B chứ còn gì

$A=\frac{2000+2001}{2001+2002}$A=2000+20012001+2002 

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

=>A=B

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

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

Ta thấy \(\frac{2000}{2001+2002}< \frac{2000}{2001}\)

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

\(\Rightarrow B< A\)

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

\(\Leftrightarrow A=\frac{2000}{2001}+\frac{2001}{2002}=\frac{2000+20001}{2001+2002}\)   VÀ  \(B=\frac{2000+2001}{2001+2002}\)

\(\Rightarrow A=B\)

chắc mk làm sai

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

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

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

Cộng vế theo vế:

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

Hay 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

mình lớp5  nhưng mình bt làm

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

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

Vậy        \(A>B\)

A=1-2000/2001=2001/2001-2000/2001=1/2001

B=1-2000/2001=2001/2001-2000/2001=1/2001

Ta thấy 1/2001=1/2001 Nên 2000/2001=2000/2001

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

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

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

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

\(\Rightarrow A>B\)

A>B nha bn