A = 35/39 

35/39 thid phải

A=7(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13)

A=7(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13)

A=7(1/3-1/13)

A=7.10/39

A=70/39

Vậy A=70/39

Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)

\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)

Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)

\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(\Rightarrow2B=1-\frac{1}{15}\)

\(\Rightarrow2B=\frac{14}{15}\)

\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)

\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)

đáp án là 59​/15

   mình chắc chắn

=22/15- 1/1.3 - 1/3.5 - 1/5.7 -.........- 1/11.13 - 1/13.15

=22/15 - (1/1.3+1/3.5+....+1/13.17)

=22/15 - 1/2(2/1.3+2/3.5.........+2/13.17)

=22/15 - 1/2(1-1/3+1/3-1/4+.............+1/13-1/17)

=22/15 - 1/2(1-1/17)

=22/15-8/17

=254/255

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)

\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{1.13}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)

\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(\Rightarrow2B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(\Rightarrow B=\frac{10}{39}:2=\frac{5}{39}\)

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

vì  ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127

=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0

=>  (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 )  x 0

= 0

b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)

= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\frac{-2}{7}-\frac{7}{18}+\frac{4}{35}\)

\(=\left(\frac{-1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-9-2-7}{18}\right)+\left(\frac{21+4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-18}{18}\right)+\left(\frac{25}{35}\right)+\frac{1}{2006}\)

\(=\left(-1\right)+\frac{5}{7}+\frac{1}{2006}\)\(=\frac{-4005}{14042}\)

b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)

\(=\left(\frac{1}{3}+\frac{1}{2007}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)\)

\(=\left(\frac{669}{2007}+\frac{1}{2007}-\frac{446}{2007}\right)-\left(\frac{27}{36}+\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)\)

\(=\frac{224}{2007}-\frac{28}{36}+\frac{10}{15}\)

\(=\frac{224}{2007}-\frac{1561}{2007}+\frac{1338}{2007}\)\(=\frac{1}{2007}\)

a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)

= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)

=  90 . 7/90 + 194/90

= 630/90 + 194/90

= 824/90 = 412/45

b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35

= -2/5 + 3/7 - 4/9 -  3/5 + 13/35 + 7/35

= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)

= -1 + 3/7 - 4/9 + 4/7

= -1 + (3/7 + 4/7) - 4/9

= -1 + 1 - 4/9 = -4/9

\(C=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

\(C=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{13}\)

\(C=1-\frac{1}{13}\)

\(C=\frac{12}{13}\)

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\frac{1}{1}-\frac{1}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(C=\frac{1}{1}-\frac{1}{13}\)

\(C=\frac{12}{13}\)

\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(C=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{11}-\frac{1}{13}\right)\)

\(C=1-\frac{1}{13}\)

\(C=\frac{12}{13}\)