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: 

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

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

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

\(\Rightarrow A>B\)

A>B nha bn

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

\(\dfrac{2000}{2001}+\dfrac{1}{2001}=1\) mà \(\dfrac{1}{2001}< \dfrac{1}{2}\) nên \(\dfrac{2000}{2001}>\dfrac{1}{2}\)

\(\dfrac{2001}{2002}+_{ }\dfrac{1}{2002}=1\) mà \(\dfrac{1}{2002}< \dfrac{1}{2}\)nên \(\dfrac{2001}{2002}>\dfrac{1}{2}\)

- Hay A>1

- mà 2000+2001<2001+2002 nên B< 1

=> 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