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

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

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

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{9\cdot11}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{9\cdot11}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{11-9}{9\cdot11}\right)\)

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

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

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

\(=\frac{1}{2}\cdot\frac{10}{11}\)

\(=\frac{10}{22}=\frac{5}{11}\)

Ta có : 

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\)\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

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

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

\(=\)\(\frac{1}{2}.\frac{10}{11}\)

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

Bạn làm đúng òi 

Chúc bạn học tốt ~

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

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

\(=\frac{1}{2}.\frac{10}{11}\)

\(=\frac{10}{22}=\frac{5}{11}\)

k nha

M=1/2{1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99}
   =1/2{1/3-1/99}
   =1/2*32/99
   =16/99

           \(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

Ta có: \(B=1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{61.63}+\dfrac{2}{63.65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}=\dfrac{133}{195}.\)

Vậy \(B=\dfrac{133}{195}.\)

a) 

\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)

b)\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)

c) Ta có: \(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\cdot\left(n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(=1-\dfrac{1}{n+1}\)

\(=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)

d)

 \(M=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\\ =\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

M=2/3.5+2/5.7+...+2/97.99

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

M=1/3-1/99

M=32/99

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

      \(=2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-...-\frac{1}{97}+\frac{1}{99}\right)\)

        \(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

          \(=2.\frac{32}{99}\)

            \(=\frac{64}{99}\)

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 2/9 +...........+ 1/97 - 1/99

= 1/3 - 1/99

= 33/99 - 1/99

= 32/99