mình lớp5  nhưng mình bt làm

Xét B=\(\frac{2000+2001}{2001+2002}\)\(=\)\(\frac{2000}{2001+2002}\)\(+\)\(\frac{2001}{2001+2002}\)

Mà  \(\frac{2000}{2001}>\frac{2000}{2001+2002}\);     \(\frac{2001}{2002}>\frac{2001}{2001+2002}\)                                                                                                  \(\Rightarrow\)\(\frac{2000}{2001}+\frac{2001}{2002}\)\(>\frac{2000+2001}{2001+2002}\)

Vậy        \(A>B\)

A=1-2000/2001=2001/2001-2000/2001=1/2001

B=1-2000/2001=2001/2001-2000/2001=1/2001

Ta thấy 1/2001=1/2001 Nên 2000/2001=2000/2001

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

\(\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

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

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

Ta thấy \(\frac{2000}{2001+2002}< \frac{2000}{2001}\)

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

\(\Rightarrow B< A\)

\(A=\frac{2000}{2001}+\frac{2001}{2002}\) VÀ\(B=\frac{2000+2001}{2001+2002}\)

\(\Leftrightarrow A=\frac{2000}{2001}+\frac{2001}{2002}=\frac{2000+20001}{2001+2002}\)   VÀ  \(B=\frac{2000+2001}{2001+2002}\)

\(\Rightarrow A=B\)

chắc mk làm sai

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

Ta có: \(\frac{2000}{2001}>\frac{2000}{2001+2002}\)

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

Cộng vế theo vế:

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

Hay A > B

a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)

Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)

hay \(\frac{3}{-4}< \frac{-1}{-4}\)

Quy đồng mẫu ta được:

15/17=15.27/17.27=405/459

25/27=25.17/27.27=425/459

⇒405/459<425/459⇒15/17<25/27

c) quy đồng

-9/6=-36/24

6/-4=36/-24

tương đương -36/24=-36/24

suy ra -9/6=6/-4

                                         Giải

Ta có\(A=\frac{2002}{2001}+\frac{2001}{2002}\)và \(B=\frac{2000}{2001}+\frac{2001}{2002}\)

Ta nhận xét thấy A và B cùng có chung 1 số hạng là \(\frac{2001}{2002}\)

Nên ta chỉ so sánh \(\frac{2002}{2001}\)và \(\frac{2000}{2001}\)ta so sánh 2 phân số đó với 1

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

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

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

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

Vậy A>B

kp nha

So sánh:

\(A=\frac{2000}{2001}+\frac{2001}{2002}\)           và           \(B=\frac{2000+2001}{2001+2002}\)

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

Ta có:

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

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

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

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

\(\Rightarrow A>B\)

Vậy: \(A>B\)

Ta có:

\(\frac{2000}{2001}\)> \(\frac{2000}{2001+2002}\)(1)

\(\frac{2001}{2002}\)> \(\frac{2001}{2001+2002}\)(2)

Cộng các bất đẳng thức (1) và ( 2) vế với nhau:

Vậy \(\frac{2000}{2001}\)+ \(\frac{2001}{2002}\)> \(\frac{2000+2001}{2001+2002}\)hay A > B.