A=2000/2001+2001/2000,B=2000/2001+2001/2002
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\frac{2000}{2001}>\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
![](https://rs.olm.vn/images/avt/0.png?1311)
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
Ta có : 2000/2001 > 2000/ 2001 + 2002 (1)
2001/2002 > 2001/2001+2002(2)
Cộng các bất đẳng thức (1) và (2) vế với nhau:
Vậy 2000/2001 + 2001/2002> 2000/2001+2002 hay A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
B= \(\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)
Ta có : \(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002};\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)
\(\Rightarrow\) \(\dfrac{2000}{2001}+\dfrac{2001}{2000}>\dfrac{2000+2001}{2001+2002}\)
\(\Rightarrow\) A>B
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