![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2000/2001 * 2002/2003 * 2001/2002 * 2003/2004*2006/2000
=((2000/2001).2002):2003.2001/2002).2003):2004.2006)/2000
=1.000998004
\(\dfrac{2000}{2001}\cdot\dfrac{2002}{2003}\cdot\dfrac{2001}{2002}\cdot\dfrac{2003}{2004}\cdot\dfrac{2006}{2000}=\dfrac{2006}{2004}=\dfrac{1003}{1002}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
phần bù đến 1 của 2000/2001 là 1- 2000/2001=1/2001
phần bù đến 1 của 2001/2002 là 1-2001/2002=1/2002
Vì 1/2001>1/2002 nên 2000/2001<2001/2002
![](https://rs.olm.vn/images/avt/0.png?1311)
2001/2000=1+1/2000
2002/2001=1+1/2001
Mà 1/2000>1/2001
=>1+1/2000>1+1/2001
hay 2001/2000>2002/2001
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
B=\(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Do \(\frac{2000}{2001}>\frac{2000}{2001+2002};\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
Ta có:$B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}$B=20002001+2002 +20012001−2002
Vì:$\frac{2000}{2001}>\frac{2000}{2001+2002}$20002001 >20002001+2002
$\frac{2001}{2002}>\frac{2001}{2001+2002}$20012002 >20012001+2002
$\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)$⇒(20002001 +20012002 )>(20002001−2002 −20012001+2001 )
$\Rightarrow A>B$⇒A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy:
2000/2001 = 1 – 1/2001
2001/2001 = 1 – 1/2002
……..
2015/2016 = 1 – 1/2016
Trong biểu thức A có 2015-2000+1=16 (số hạng). Nên
A = 2000/2001 + 2001/2002 + .....+ 2015/2016
A = 16 – (1/2001 + 1/2002 + 1/2003 + … + 1/2016) (1)
Mà:
1/2001 ; 1/2002 ; 1/2003 ; … ; 1/2016 đều bé hơn 1/2000. Nên:
1/2001 + 1/2002 + 1/2003 + … + 1/2016 < 16/2000 < 1 (2)
Từ (1) và (2) suy ra:
A = 16 – (1/2001 + 1/2002 + 1/2003 + … + 1/2016) > 15
A > 15
Ta thấy :
2000/2001 = 1 - 1/2001
2001/2002 = 1 - 1/2002
.................................
2015/2016 = 1 - 1/2016
Trong biểu thức A có :
2015 - 2000 + 1 = 16 ( số hạng )
A = 2000/2001 + 2001/2002 + .... + 2015/2016
A 16 - ( 1/2001 + 1/2002 + 1/2003 + ... = 1/2016 ) ( 1 )
Mà :
1/2001 ; 1/2002 ; 1/2003 ;...;1/2016 đều bé hơn 1/2000 Nên
1/2001 + 1/2002 + 1/2003 + ... + 1/2016 < 16/2000 < 1 ( 2 )
Từ ( 1 ) và ( 2 ) suy ra :
A = 16 - ( 1/2001 + 1/2002 + 1/2003 + ... +1/2016 ) < 15
A > 15
\(\frac{2001}{2000}-\frac{2002}{2001}=\frac{2001.2001}{2000.2001}-\frac{2002.2000}{2000.2001}\)
\(=\frac{\left(2002-1\right).\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.\left(2000+1\right)-\left(2000+1\right)-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+2002-2000-1-2002.2000}{2000.2001}\)
\(=\frac{2002.2000+1-2002.2000}{2000.2001}\)
\(=\frac{1}{2000.2001}\)