tôi nghĩ đề thế này đúng hơn

A=2000/2001+2001/2000,B=2000+2001/2001+2002

hoặc ngược lại

ta thấy

2000/2001=2000/2001

2001/2000=2001/2002

=>2000/2001+2001/2000=2000/2001+2001/2002

=>A=B

mk nghĩ đề sai vì lớp 6 sao lại có kiểu so sánh quá dễ như vậy

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

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

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

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

Đề sai chỗ 2001/2001 phải là 2001/2002

\(A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2002}+\dfrac{2001}{2002}=\dfrac{4001}{2002}>1\)

B=\(\dfrac{2000+2001}{2001+2002}=\dfrac{4001}{4003}< 1\)

=>A>B

A > B

Ta có :

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

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

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

Nên \(\dfrac{2000}{2001}+\dfrac{2001}{2002}\) \(>\dfrac{2000+2001}{2001+2002}\)

Vậy A > B

B=2000+1+2002=4003

A=2000/2001+2001/2002

=2002.(2000+2001)/2001.2002

=2000+2001/2001<1

Mà B>1 suy ra A<B