A=B (do 2 phân số giống nhau)

Ta có: \(A=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)và\(B=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)

Mà\(\dfrac{2000}{2001}+\dfrac{2001}{2002}=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)

Vậy A=B

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

Do đó: A=B

Trong phần câu hỏi tương tự có đó!                                                    

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

Ta có:

2000/2001 > 2000/2001+2002

2001/2002 > 2001/2001+2002

Vậy A >B

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

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

Ta có:

2000/2001 > 2000/2001+2002

2001/2002 > 2001/2001+2002

Vậy A >B

Ta có 

B= 2000/2001+2002 + 2001/2001+2002.                                                           Mà 2000/2001+2002 < 2000/2001 và 2001/2001+2002 < 2001/2002.              Nên 2000/2001+2002 + 2001/ 2001+2002 < 2000/2001 + 2001/2002.       Hay 2000+2001/ 2001+2002 < 2000/2001 + 2001/2002                             Suy ra B < A

A>B

CÁC BẠN NHỚ **** CHO MÌNH NHA

ta có

B = 2000/20001 + 20002 + 2001/2001+2002

mà 2000/2001+2002<2000/2001

và 2001/2001+2002<2001/2002

nên 2000+2001/2001+2002<2000/2001+2001/2002

hay 2000+2001/2001+2002<2000/2001+2001/2002

ta có:\(A=\frac{2000}{2001}+\frac{2001}{2002}<\frac{2000}{2002}+\frac{2001}{2002}=\frac{2000+2001}{2002}<\frac{2000+2001}{2001+2002}=B\)

\(\Rightarrow A

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

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

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

=>A>B

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

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