S=-4/1.4/5-4/5.4/9-4/9.4/13-...-4/2013.4/2017

=-4^2/1.5-4^2/5.9-4^2/9.13-...-4^2/2013.2017

=-4(4/1.5+4/5.9-4/9.13-...-4/2013.2017)

=-4(1-1/5+1/5-1/9+1/9-1/13-...-1/2013-1/2017)

=-4(1-1/2017)

=-4.2016/2017=-8064/2017

vậy 2017S=-8064

Ta có : S= - 4/1 .4/5 - 4/5 .4/9-...- 4/2013 .4/2017

             = -4 .( 4/1.5+4/5.9+...+4/2013.2017 )

            = -4 . (1-1/5+1/5-1/9+...+1/2013-1/2017 )

           = -4 . (1-1/2017)

           = -4 .2016/2017

           = -8064/2017 

=> 2017S= -8064/2017 .2017 = -8064

\(\left(-1\frac{1}{6}\right)\left(\frac{1-\frac{3}{5}+\frac{3}{11}-\frac{3}{13}}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right)\left(\frac{4-\frac{4}{17}+\frac{4}{19}-\frac{4}{2013}}{5-\frac{5}{7}+\frac{5}{19}-\frac{5}{2013}}\right)\)

\(=-\frac{7}{6}.\left(\frac{3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}\right)}{\frac{1}{3}-\frac{1}{5}+\frac{1}{11}-\frac{1}{13}}\right):\left(\frac{4.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}{5.\left(1-\frac{1}{7}+\frac{1}{19}-\frac{1}{2013}\right)}\right)\)

\(=-\frac{7}{6}.3:\frac{4}{5}=-\frac{7}{2}.\frac{5}{4}=-\frac{35}{8}\)

1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

= \(\frac{5}{9}\) .(\(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\) )

= \(\frac{5}{9}\) . 1 = \(\frac{5}{9}\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)