+ \(\frac{2000}{2001}=\frac{2001-1}{2001}=1-\frac{1}{2001}\)

+ \(\frac{2001}{2002}=\frac{2002-1}{2002}=1-\frac{1}{2002}\)

+ \(\frac{1}{2001}>\frac{1}{2002}\Rightarrow1-\frac{1}{2001}

\(1-\frac{2000}{2001}=\frac{1}{2001}\)

\(1-\frac{2001}{2002}=\frac{1}{2002}\)

Vì \(\frac{1}{2001}>\frac{1}{2002}\) nên  \(\frac{2000}{2001}

Quá dễ

Ta có: 2000/2001 = 1 - 1/2001

          2001/2002 = 1 - 1/2002

mà 1/2001 > 1/2002

  --> 1 - 1/2001 < 1 - 1/2002

-->      2000/2001 < 2001/2002

Ta thấy: \(\frac{2000}{2001}=1-\frac{1}{2001}\)

              \(\frac{2001}{2002}=1-\frac{1}{2002}\)

Vì: \(\frac{1}{2001}>\frac{1}{2002}\)

\(\Rightarrow\frac{2000}{2001}< \frac{2001}{2002}\)

Ta có:1-2000/2001=1/2001                                              1-2001/2002=1/2002

vì 1/2001>1/2002.Suy ra 2000/2001<2001/2002

\(\frac{2000}{2001}=1-\frac{1}{2001}\)

\(\frac{2001}{2002}=1-\frac{1}{2002}\)

\(2001< 2002\Rightarrow\frac{1}{2001}>\frac{1}{2001}\)

                             \(\Rightarrow1-\frac{1}{2001}< 1-\frac{1}{2002}\)

                              \(\Rightarrow\frac{2000}{2001}< \frac{2001}{2002}\)

ta có:2000/2001=1-1/2001

2001/2002=1-1/2002

mà 2001<2002

suy ra 1/2001>1/2002

suy ra 1-1/2001<1-1/2002

vậy 2000/2001<2001/2002

Trả lời:

Ta có: \(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)

          \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2001}-\frac{1}{2002}=1-\frac{1}{2002}\)

Ta thấy: \(2001< 2002\)

\(\Leftrightarrow\frac{1}{2001}>\frac{1}{2002}\)

\(\Leftrightarrow-\frac{1}{2001}< -\frac{1}{2002}\)

\(\Leftrightarrow1-\frac{1}{2001}< 1-\frac{1}{2002}\)

\(\Leftrightarrow\frac{2000}{2001}< \frac{2001}{2002}\)

Vậy ....

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

 13/27 và 7/15
\(\frac{13}{27}\) = 1:\(\frac{27}{13}\)= 1: \(\frac{26+1}{13}\) = 1: ( 2+\(\frac{1}{13}\))
\(\frac{7}{15}\)= 1:\(\frac{15}{7}\)= 1: \(\frac{14+1}{7}\)= 1: ( 2+ \(\frac{1}{7}\))
ta có \(\frac{1}{13}\)< \(\frac{1}{7}\)=>   2+\(\frac{1}{13}\)< 2+ \(\frac{1}{7}\) => 1: ( 2+\(\frac{1}{13}\)) >  1: ( 2+ \(\frac{1}{7}\))

vậy \(\frac{13}{27}\)>\(\frac{7}{15}\)

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

Ta có \(\frac{2000}{2001}=1-\frac{1}{2001}\)

          \(\frac{2001}{2002}=1-\frac{1}{2002}\)

Vì \(\frac{1}{2001}>\frac{1}{2002}\)

=> \(1-\frac{1}{2001}< 1-\frac{1}{2002}\)

=> \(\frac{2000}{2001}< \frac{2001}{2002}\)

ta thấy                                                                                                                                                                                      \(1=\frac{2000}{2001}+\frac{1}{2001}\)

\(1=\frac{2001}{2002}+\frac{1}{2002}\)

  mà \(\frac{1}{2001}\) \(>\frac{1}{2002}\)   ( phần bù )

   \(\frac{\Rightarrow2000}{2001}< \frac{2001}{2002}\)

2001/2002 lớn hơn

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

1- \(\frac{2000}{2001}\)= \(\frac{1}{2001}\)

1 - \(\frac{2001}{2002}\)=  \(\frac{1}{2002}\)

Vì \(\frac{1}{2001}\)< \(\frac{1}{2002}\)nên \(\frac{2000}{2001}\)< \(\frac{2001}{2002}\)

2001/2000 > 2002/2001

j

o

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

Ta có : 

1 - 2000/2001 = 1/2001 

1 - 2001/2002 = 1/2002 

Vì 1/2001 > 1/2002 nên 2000/2001 > 2001/2002