\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{4}{15}+\frac{9}{5}\)

\(=\frac{31}{15}\)

              Bài làm :

Ta có :

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}+\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{9\times10}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{15}+2\left(1-\frac{1}{10}\right)\)

\(=\frac{31}{15}\)

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

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

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

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

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

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

\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

\(y=\frac{2}{3}.\frac{11}{10}\)

\(y=\frac{22}{30}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

         \(\frac{10}{11}.y=\frac{2}{3}\)

                    \(y=\frac{11}{15}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{11-9}{9.11}\right).y=\frac{2}{3}\)

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

\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

         \(y=\frac{2}{3}:\frac{10}{11}\)

         \(y=\frac{11}{15}\)

a ) A = 20,15 x 25,75 + 74,25 x 20,15

     A = 20,15 x ( 25,75 + 74,25 )

     A = 20,15 x       100

     A =     2015

Tính bằng cách thuận tiện nhất 

a) A = 20,15 x 25,75 + 74,25 x 20,15

= 20,15 x (25,75 + 74,25)

= 20,15 x      100

= 2015

 b ) B = 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 + 2/9x7

      B = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9

       B =  1 + ( 1/3 - 1/3 ) + ( 1/5 - 1/5 ) + ( 1/7 - 1/7 ) - 1/9

       B =  1 - 1/9

       B =   8/9

bạn có thể tìm kiếm , có nhiều người hỏi phàn B lắm

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

\(S.2=\frac{1}{1}-\frac{1}{11}\)

\(S.2=\frac{10}{11}\)

\(S=\frac{10}{11}:2\)

\(S=\frac{5}{11}\)

S = 5/11

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

\(=\frac{1}{1}-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{2}{1x3}+\)\(\frac{2}{3x5}+\)\(\frac{2}{5x7}+\)\(\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}\)

= \(\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\)\(+\frac{13-11}{11x13}\)

= \(\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+\frac{9}{7x9}-\frac{7}{7x9}+\frac{11}{9x11}\)\(-\frac{9}{9x11}\)\(+\frac{13}{11x13}-\frac{11}{11x13}\)

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

= \(1-\frac{1}{13}=\frac{12}{13}\)

12/13

Bạn làm ra được ko?

= 1-1/3 + 1/3-1/5+.......+1/97-1/99

=  1 - 1/99

= 98/99

sao lại là 1- 1/3 + 1/3 -1/5 + ...... 1/97 - 1/99 hả bạn :|

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{95.97}\)\(+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

M=1/3-1/5+1/5-1/7+.....+1/97-1/99

M=1/3-1/199

M=32/99

\(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(M=\frac{1}{3}-\frac{1}{99}\)

\(M=\frac{32}{99}\)

~Học tốt~

\(M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

     \(=\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

     \(=\)\(1-\frac{1}{99}\)

     \(=\)\(\frac{98}{99}\)

vậy \(M=\frac{98}{99}\)