Ta có: \(\frac{2004\cdot2007+6}{2005\cdot2005+2009}=\frac{\left(2005-1\right)\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-1\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-2007+6}{2005\cdot2005+2009}\)

\(=\frac{\text{2005 x (2005 + 2) - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 2005 x 2 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 4010 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\text{ }\frac{\text{2005 x 2005 + 2009}}{\text{2005 x 2005 + 2009}}=1\)

Giúp mình nha

Ta có: 1/ 2001 . 2003 = 1/2001 - 1/2003...

=> 1/2001 - 1/2003 + 1/2003 - 1/2005 + 1/2005 - 1/2007 + ... +1/2009 - 1/2011 +1/2011 - 1/2013

= 1/2001 - 1/2013

= 4/ 1342671

\(\frac{1}{2001\times2003}\)+\(\frac{1}{2003\times2005}\)+\(\frac{1}{2005\times2007}\)+........+\(\frac{1}{2009\times2011}\)+\(\frac{1}{2011\times2013}\)

=\(\frac{1}{2001}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2007}\)+........+\(\frac{1}{2009}\)-\(\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2013}\)

=\(\frac{1}{2001}\)-\(\frac{1}{2013}\)

=\(\frac{2013}{4028013}-\frac{2001}{4028013}\)=\(\frac{2}{4028013}\)

\(\frac{2005\cdot2004-1}{2003\cdot2005+2004}\)

\(=\frac{2005\cdot\left(2003+1\right)-1}{2003\cdot2005+2004}\)

\(=\frac{2005\cdot2003+2005-1}{2003\cdot2005+2004}\)

\(=\frac{2005\cdot2003+2004}{2003\cdot2005+2004}\)

\(=1\)

2005 x 2004 - 1 / 2003 × 2005 + 2004

= 2005 × (2003 + 1) - 1 / 2003 × 2005 + 2004

= 2005 × 2003 + (2005 - 1) / 2003 × 2005 + 2004

= 2005 × 2003 + 2004 / 2003 × 2005 + 2004

= 1

\(\frac{2005\times2004-1}{2003\times2005+2004}\)

\(=\frac{2005\times\left(2003+1\right)-1}{2003\times2005+2004}\)

\(=\frac{2005\times2003+\left(2005-1\right)}{2003\times2005+2004}\)

\(=\frac{2005\times2003+2004}{2003\times2005+2004}\)

\(\frac{2005x2004-1}{2003x2005+2004}\)=\(\frac{4018019}{4018019}\)= 1

                       Bài giải

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

\(\frac{2005\times2004-1}{2003\times2005+2004}\)\(=\)\(\frac{2005\times\left(2003+1\right)-1}{2003\times2005+2004}\)\(=\)\(\frac{2005\times2003+2005-1}{2003\times2005+20042}\)\(=\)\(\frac{2005\times2003+2004}{2003\times2005+2004}\)\(=\)1

a. kết quả = 401/402

b. Ta có: 1-2004/2009=5/2009     ,        1--2005/2010=5/2010 . Vì 5/2009  > 5/2010 nên 2004/2009 < 2005/2010.

Đấy phần b. mk ko quy đồng nha! 

Nhớ Tích cho mk đấy

thế phần a chỉ cần làm như vậy thôi à pạn

Alo tuổi lìn sánh vai

`2007/2009×2002/2005×2009/2006×2005/2007×2006/2002`
`=(2007xx2002xx2009xx2005xx2006)/(2009xx2005xx2006xx2007xx2002)`
`=(2007xx2002xx2009xx2005xx2006)/(2007xx2002xx2009xx2005xx2006)`
`=1`

\(\dfrac{2007}{2009}.\dfrac{2002}{2005}.\dfrac{2009}{2006}.\dfrac{2005}{2007}.\dfrac{2006}{2002}\\ =\left(\dfrac{2007}{2009}.\dfrac{2009}{2006}\right).\left(\dfrac{2006}{2002}.\dfrac{2002}{2005}\right).\dfrac{2005}{2007}\\ =\dfrac{2007}{2006}.\dfrac{2006}{2005}.\dfrac{2005}{2007}=1\)

\(\dfrac{2007}{2009}\text{×}\dfrac{2002}{2005}\text{×}\dfrac{2009}{2006}\text{×}\dfrac{2005}{2007}\text{×}\dfrac{2006}{2002}\\ =\dfrac{2007\text{×}2002\text{×}2009\text{×}2005\text{×}2006}{2009\text{×}2005\text{×}2006\text{×}2007\text{×}2002}=1\)

Áp dụng \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n+1}\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)