sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{2}{11}\)

\(=1-\dfrac{2}{11}\)

\(=\dfrac{9}{11}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(\Rightarrow B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow B=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

_Học tốt_

100/101

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+....+\frac{2}{99\times101}\)

\(=\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+....+\frac{101-99}{99\times101}\)

\(=\frac{3}{1\times3}-\frac{1}{1\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+....+\frac{101}{99\times101}-\frac{99}{99\times101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

SOS

hép☹

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{19\times21}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

                                                    \(=1-\frac{1}{21}=\frac{20}{21}\)

đúng cái nhé

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

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

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

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

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

\(=\frac{20}{11}\)