2004.2006-2003.2005

= ( 2004 - 2003 ) x ( 2006 - 2005 )

=   1                  x    1

=1

\(\dfrac{8018}{2004.2006-2003.2005}\)

\(=\dfrac{8018}{\left(2005-1\right)\left(2005+1\right)-\left(2004-1\right)\left(2004+1\right)}\)

\(=\dfrac{8018}{2005^2-1^2-2004^2+1^2}=\dfrac{8018}{\left(2005-2004\right)\left(2005+2004\right)}\)

\(=\dfrac{8018}{1.4009}=2\)

Chúc bạn học tốt!!!

Đặt:

\(HANG=\dfrac{8018}{2004.2006-2003.2005}\)

\(HANG=\dfrac{8018}{\left(2005-1\right)\left(2005+1\right)-\left(2004-1\right)\left(2004+1\right)}\)

\(HANG=\dfrac{8018}{2005^2-1-2004^2+1}\)

\(HANG=\dfrac{8018}{2005^2-2004^2}\)

\(HANG=\dfrac{8018}{\left(2005-2004\right)\left(2005+2004\right)}\)

\(HANG=\dfrac{8018}{4009}=2\)

Vậy \(HANG=2\)

\(\dfrac{8018}{2004.2006-2003.2005}\)

\(=\dfrac{8018}{\left(2005-1\right)\left(2005+1\right)-\left(2004-1\right)\left(2004+1\right)}\)

\(=\dfrac{8018}{2005^2-1^2-2004^2+1^2}\)

\(=\dfrac{8018}{\left(2005-2004\right)\left(2005+2004\right)}\)

\(=\dfrac{8018}{1.4009}=2\)

\(\frac{2004.2006+1000}{2005.2006-1006}\)

\(=\frac{2004.2006+1000}{\left(2004+1\right).2006-1006}\)

\(=\frac{2004.2006+1000}{2004.2006+2006-1006}\)

\(=\frac{2004.2006+1000}{2004.2006+1000}\)

\(=1\)

\(\frac{2004x2006+1000}{2005x2006-1006}\)

=\(\frac{1000}{1x2006-1006}\)cùng giảm 2004 lần số 2006

=\(\frac{1000}{1000}=1\)

\(\frac{2005.2004-1}{2003.2005+2004}=\frac{2005.\left(2003+1\right)-1}{2003.2005+2004}=\frac{2005.2003+2005-1}{2005.2003+2004}=\frac{2005.2003+2004}{2005.2003+2004}=1\)

A = \(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3.3^{19}-7.2^{29}.3^{18}}=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}=2\)

B = \(\dfrac{8020}{2004.2006-2003.2005}\)

Đặt x = 2004, ta có:

\(\dfrac{4x+2}{x\left(x+2\right)-\left(x-1\right)\left(x+1\right)}=\dfrac{4x+2}{2x+1}=\dfrac{2\left(2x+1\right)}{2x+1}=2\)

Ta có: 

A =  \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\) + ... + \(\frac{2}{2001.2003}\) + \(\frac{2}{2003.2005}\)

    =  \(\frac{1}{1}\) -  \(\frac{1}{3}\)+ \(\frac{1}{3}\)-  \(\frac{1}{5}\)+ \(\frac{1}{5}\)-  \(\frac{1}{7}\)+  \(\frac{1}{7}\)-  \(\frac{1}{9}\) + ... + \(\frac{1}{2001}\)- \(\frac{1}{2003}\)+ \(\frac{1}{2003}\)-  \(\frac{1}{2005}\)

    = 1 - \(\frac{1}{2005}\)

     = \(\frac{2004}{2005}\)

Chúc bạn học tốt nha ^^!!

=>2B=2/1.3 +2/3.5 +2/5.7+...+2/2003.2005

=>2B=1-1/3+1/3-1/5+1/5-1/7+...+1/2003-1/2005

=>2B=-1/2005

=>B=-1/2005:2=-1/4010

Vậy B= -1/4010

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(2A=2.\left(\frac{1}{1.3}+\frac{1}{2.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\right)\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}:2\)

\(A=\frac{1002}{2005}\)

Ủng hộ tk Đúng nha mọi người !!! ^^ 

Đặt B = \(\frac{1}{1.3}\)+ \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\Rightarrow2B=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\right)\)\(\Rightarrow2B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2003.2005}\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{2005}=\frac{2012}{6015}\Rightarrow B=\frac{2012}{6015}:2=\frac{1001}{6015}\)

Ta có: \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{2003.2005}\)

             \(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{2003.2005}\right)\)

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

               \(=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

                 \(=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)

Cho tổng trên là A

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(A=2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(A=2\left(1-\frac{1}{2005}\right)\)

\(A=2.\frac{2004}{2005}\)

\(A=\frac{4008}{2005}\)