a) ta thay 1-2002/2003= 1/2003 va 1-2003/2004=1/2004 

ma 1/2003>1/2004 =>2002/2003<2003/2004

b) ta co -2002/2003<1<2005/2004

a.2002/2003<2003/2004

b.-2002/2003<2005/-2004

neu dung thi ?

a.<;b.<

=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8

\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)

Ta có:

2002/2001=1+1/2001

2003/2002=1+1/2002

2004/2003= 1+ 1/2003

2005/2004= 1+ 1/2004

2006/2005=1+ 1/2005

2007/2006= 1+ 1/2006

2008/2007=1 + 1/2007.

2009/2008=1+ 1/2008.

=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.

Nhớ k đúng cho mình nha!! Thanks!!!

ta có 

\(-\frac{2002}{2003}>-1>\frac{2005}{-2004}.\)

\(\Rightarrow-\frac{2002}{2003}>\frac{2005}{-2004}\)

\(\frac{2002}{2003}\)>\(\frac{2005}{-2004}\)

a) Ta có: \(1-\frac{2002}{2003}=\frac{1}{2003}\)

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

Vì \(\frac{1}{2003}>\frac{1}{2004}\)

\(\Rightarrow\frac{2002}{2003}>\frac{2003}{2004}\)

b) Ta có: \(\frac{-2005}{-2004}=\frac{2005}{2004}>1\)

\(\frac{-2002}{2003}

mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15

Ta có :

\(\frac{-2002}{2003}>-1\)

\(-1>\frac{-2005}{2004}\)

\(\Rightarrow\frac{-2002}{2003}>\frac{-2005}{2004}\)

Ta có: 

\(-\frac{2002}{2003}>-1\)

\(-\frac{2005}{2004}< -1\)

=> \(-\frac{2002}{2003}>-\frac{2005}{2004}\)

Ta có :

\(\frac{-2002}{2003}>-1\)

\(\frac{-2005}{2004}< -1\)

\(\Rightarrow\frac{-2002}{2003}>\frac{-2005}{2004}\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)

Ta có:

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)

\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)

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

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!

                                 \(\text{ Bài giải}\)

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

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

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

        \(\text{Vậy : }A>B\)