\(B=\frac{2003+2004}{2004+2005}=\frac{2003}{2004+2005}+\frac{2004}{2004+2005}\)

Ta có: \(\frac{2003}{2004}>\frac{2003}{2004+2005}\)

          \(\frac{2004}{2005}>\frac{2004}{2004+2005}\)

\(\frac{2003}{2004}+\frac{2004}{2005}>\frac{2003+2004}{2004+2005}\)

\(A>B\)

Vậy A>B

đáp án là A>B

chúc bn hok tốt!

                                 \(\text{ Bài giải}\)

\(A=\frac{2003}{2004}+\frac{2004}{2005}=0,999500998 + 0,999501247=1.99900225\)

\(B=\frac{2003+2004}{2004+2005}=\frac{4007}{4009}=0,999501122\)

\(\text{Vì : }1,99900224>0,999501122\text{ nên }A>B\)

        \(\text{Vậy : }A>B\)

Ta có:
N=\(\dfrac{2003+2004}{2004+2005}\)=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Ta thấy:
\(\dfrac{2003}{2004+2005}\)<\(\dfrac{2003}{2004}\)(1)

\(\dfrac{2004}{2004+2005}\)<\(\dfrac{2004}{2005}\)(2)
Từ (1) và (2) --> M=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\)>\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)=N
Vậy  M>N

đáp án là:A>B

chúc bn hok tốt

\(A=\frac{20032}{2004}+\frac{2004}{2005}=9,99600798+0,999501247=10,9955092\)

\(B=\frac{2003+2004}{2004+2005}=\frac{4007}{4009}\)

\(\text{Vì : }10,9955092>1\text{ mà }\frac{4007}{4009}< 1\text{ nên }10,9955092>\frac{4007}{4009}\)

\(\text{Vậy : }A>B\)

\(2004A=\frac{2004^{2004}+2004}{2004^{2004}+1}=1+\frac{2003}{2004^{2004}+1}\)

\(2004B=\frac{2004^{2005}+2004}{2004^{2005}+1}=1+\frac{2003}{2004^{2005}+1}\)

\(\frac{2003}{2004^{2004}+1}>\frac{2003}{2004^{2005}+1}\)

\(\Rightarrow2004A>2004B\)

\(\Rightarrow A>B\)

2004A=\(\frac{2004^{2004}+2004}{2004^{2004}+1}\)

\(\frac{2004^{2004}+2004}{2004^{2004}+1}-1=\frac{2003}{2004^{2004}+1}\)

2004B=\(\frac{2004^{2005}+2004}{2004^{2005}+1}\)

\(\frac{2004^{2005}+2004}{2004^{2005}+1}-1=\frac{2003}{2004^{2005}+1}\)

Ta thấy :\(\frac{2003}{2004^{2004}+1}>\frac{2003}{2004^{2005}+1}\)

=> \(2004A>2004B\)

Vậy \(A>B\)

Ta thấy :

2004/2005 >2004/2005+2006

2005/2006> 2005/2005+2006

=> 2004/2005 + 2005/2006 >2004+2005 / 2005+2006

Ta có: \(\frac{2004}{2005}>\frac{2004}{2005+2006}\) (1)

          \(\frac{2005}{2006}>\frac{2005}{2005+2006}\) (2)

Từ (1) và (2) => \(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004+2005}{2005+2006}\) => M>N

Ai k mik mik k lại. chúc các bạn thi tốt

Bạn tham khảo nhé 

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\) \(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{2004^{2004}+1}{2004^{2005}+1}< \frac{2004^{2004}+1+2003}{2004^{2005}+1+2003}=\frac{2004^{2004}+2004}{2004^{2005}+2004}=\frac{2004\left(2004^{2003}+1\right)}{2004\left(2004^{2004}+1\right)}=\frac{2004^{2003}+1}{2004^{2004}+1}\)

Lại có : 

\(A=\frac{2004^{2003}+1}{2004^{2004}+1}\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vậy \(A>B\)

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