2006 x 2005 - 1/2004 x 2006 + 2005

= 2006 x ( 2004 + 1 ) - 1/2004 x 2006 + 2005

= 2006 x 2004 + 2006 x 2005/2004 x 2006 + 2005

= 2006 x 2004 + 2005 / 2004 x 2006 + 2005

dung roi ban cam on ban nhe

ừ ko có gì

2010

tick mình nha 

5 - ( 1997 -2005)+1997

=5 -(-8)+1997

=13+1997

=2010

đề bài bạn ghi sai rồi ko phải ''= 1997'' mà là ''+ 1997''

\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)

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

Do \(\frac{1}{2003}>\frac{1}{2004}>\frac{1}{2005}.\) nên \(\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>0\)

Vì vậy \(3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>3\) (đpcm)

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

\(=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})\)

\(=3+(\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005})\)

Do\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)

\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)

\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)

\(\Rightarrow A\)>\(3\)

A=\(\frac{2006x\left(2004+1\right)-1}{2006x2004+2005}=\frac{2006x2004+2006-1}{2006x2004+2005}\)

= \(\frac{2006x2004+2005}{2006x2004+2005}=1\)

\(\frac{2006\times2005-1}{2004\times2006+2005}=\frac{2006\times\left(2004+1\right)-1}{2004\times2006+2005}=\frac{2006\times2004+2006\times1-1}{2004\times2006+2005}=\frac{2006\times2004+2005}{2006\times2004+2005}=1\)

\(\frac{2006\times2005-1}{2004\times2006+2005}=\frac{2006\times\left(2004+1\right)-1}{2004\times2006+2005}=\frac{2004\times2006+2006\times1-1}{2004\times2006+2005}\)

\(=\frac{2004\times2006+2005}{2004\times2006+2005}=1\)

\(\frac{2006x2005-1}{2004x2006+2005}\)

\(=\frac{2006x\left(2004+1\right)-1}{2004x2006+2005}\)

\(=\frac{2006x2004+2006-1}{2004x2006+2005}\)

\(=\frac{2006x2004+2005}{2004x2006+2005}\)

\(=1\)

 (2007 – 2005) + (2003 – 2001) +...+ (7 – 5) + (3 – 1)

= 2 + 2 + ... + 2 + 2 

= 2.502

= 1004

là cái        :D

giải:

dãy số trên có các cặp là:

[(2007-1) : 2 + 1] : 2=502

(2007-2005)+(2003-2001)+.........+(7-5)+(3-1)

=     2          +        2          +...........+ 2   + 2

=     2 . 502( dấu chấm này là dấu nhân nhá bạn)

=        1004

vậy tổng của dãy số trên là 1004