\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

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

\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)

 \(A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

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

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

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

ta có A=1-1/3+1/2-1/5+..................1/95-1/97+1/97-1/99

        A=1-1/99

        A=98/99

Cho A =2/1.3+2/3.5+2/5.7+2/7.9+....2/97.99 

A=1-1/3+1/3-1/5+1/5-1/7+..........+1/97-1/98

A=1-1/98

A=98/99

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}\)

\(=\dfrac{2021}{2022}\)

2/2*[2/1-2/2022]=2021/1011

Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)

\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{n\cdot\left(n+2\right)}<\frac{2003}{2004}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}<\frac{2003}{2004}\)

\(\Rightarrow1-\frac{1}{n+2}<\frac{2003}{2004}\)

\(\Rightarrow\frac{1}{n+2}>\frac{1}{2004}\)

\(\Rightarrow n+2<2004\)

\(\Rightarrow n=2002\)

nhầm bước cuối

\(\Rightarrow n<2002\)

giúp mình với

\(\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right].y=\frac{2}{3}\)

\(\Leftrightarrow\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right].y=\frac{2}{3}\)

\(\Leftrightarrow\left[\frac{1}{1}-\frac{1}{9}\right].y=\frac{2}{3}\)

\(\Leftrightarrow\frac{8}{9}.y=\frac{2}{3}\)

\(\Leftrightarrow y=\frac{2}{3}:\frac{8}{9}\)

\(\Leftrightarrow y=\frac{3}{4}\)

( \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\))                                    x \(y\) = \(\frac{2}{3}\)

( \(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\))                                      x \(y\) = \(\frac{4}{3}\)               (nhân 2 vế lên với 2)

(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)- \(\frac{1}{11}\))         x     \(y\)= \(\frac{4}{3}\)

( 1 - \(\frac{1}{11}\))                                                                        x    \(y\)=\(\frac{4}{3}\)

\(\frac{10}{11}\)                  x            \(y\)                                                       =\(\frac{4}{3}\)

                                              \(y\)                                                      = \(\frac{4}{3}\): \(\frac{10}{11}\)

                                              \(y\)                                                       = \(\frac{4}{3}\)x \(\frac{11}{10}\)

                                               \(y\)                                                       =\(\frac{22}{15}\)

kết quả đúng nhưng mình ko hiểu bạn có thể giáng lại ko ?

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