Ta có: A = \(\frac{6}{5\times7}+\frac{6}{7\times9}+\frac{6}{9\times11}+...+\frac{6}{95\times97}+\frac{6}{97\times99}\)

\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{95\times97}+\frac{1}{97\times99}\right)\)

\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{99}\right)\)

=> A = ...

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

b)

S2=6/2x5+6/5x8+6/8x11+...+6/29x32

=2.(3/2.5+3/5.8+...+3/29.32)

=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2.(1/2-1/32)

=2.15/32

=15/16

 Đúng 1 

a)

Ta có:

S1=2/3x5+2/5x7+2/7x9+...+2/97x99

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

=1/3-1/99

=32/99

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)

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

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

\(=1-\frac{1}{9}=\frac{8}{9}\)

 ~ Hok tốt ~

mình vẫn không hiểu

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 \(M=\frac{98}{99}\)

= 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 :|

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

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

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

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

\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)

\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)=3\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=3\cdot\dfrac{56}{305}=\dfrac{168}{305}\)