1, \(\dfrac{16\times25-22\times16}{7\times3+5\times7}=\dfrac{16\times\left(25-22\right)}{7\times\left(5+3\right)}=\dfrac{16\times3}{7\times8}\)

\(=\dfrac{6}{7}\)

2,\(\dfrac{2001\times2003+2003\times2005}{2003\times4006}=\dfrac{2003\times\left(2001+2005\right)}{2003\times4006}=\dfrac{2003\times4006}{2003\times4006}=1\)

\(\dfrac{1}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)=\dfrac{2}{1342671}\)

=1/2.(2/2001.2003+2/2003.2005+.....+2/2011.2013)

=1/2(1/2001-1/2003+1/2003-1/2005+....+1/2011-1/2013)

=1/2(1/2001-1/2013)

=2/1342671

a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

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

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005

2005 x 2004 - 1/2003 x 2005 + 2004

=2004 x ( 2005 + 1 ) - ...?

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

= 2005 x 2003 + 2005 x 1 - 1/ 2003 x 2005 + 2004

= 2005 x 2003 + 2005 -1/ 2003 x 2005 + 2004

= 2003 x 2005 + 2004/ 2003 x 2005 + 2004

= 1 ( vì TS=MS)

Theo cách mk học sẽ suy ra lun

=1/1-1/3+1/3-1/5+1/5-1/7+...+1/2001-1/2003+1/2003-1/2005

=1-1/2005

=2004/2005

Tớ nghĩ là .................................................botay.kom

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

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

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

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

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

=1

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

= 2005 x 2003 + 2005 x 1 - 1/ 2003 x 2005 + 2004

= 2005 x 2003 + 2005 -1/ 2003 x 2005 + 2004

= 2003 x 2005 + 2004/ 2003 x 2005 +2004