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

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

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

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

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

Ta có : \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{11.13}\)

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

\(=\frac{1}{3}-\frac{1}{13}\)

\(=\frac{10}{39}\)

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\)

\(=\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\)

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

\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

( bn tách ra mỗi phân số thành 1 hiệu nhé) ta có2/3.5=2/3-2/5

2/5.7=2/5-2/7

.......=........

2/11.13=2/11-2/13

=>2/3.5+2/5.7+...+2/11.13=2/3-2/5+2/5-2/7+....+2/11-2/13=2/3-2/13=..... ( bn tự tính nha)

(Dạng này bạn nên tìm hiểu và làm nhìn bt nhé,tên" tính tổng dãy ps nâng cao" nhak,mik giải như thế hơi khó hỉu,bn chịu khó Vy lại ra giấy nhak^_^ chúc học tốt và làm bài đúng nhak)

Ta có: \(\frac{2}{3\times5}+\frac{2}{5\times7}+\cdots+\frac{2}{49\times51}\)

\(=\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{49}-\frac{1}{51}\)

\(=\frac13-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)

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

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=1-\dfrac{6}{21}=\dfrac{15}{21}=\dfrac{5}{7}\)

M=\(\frac{2}{3\times5}+\frac{2}{5\times7}+.............+\frac{2}{97\times99}\)

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

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

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

\(\Leftrightarrow M=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{97}-\frac{2}{99}\)

\(\Rightarrow M=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

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

\(\Rightarrow M=2\times\frac{32}{99}\)

\(\Rightarrow M=\frac{64}{99}\)

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

=1/3-1/99

=32/99

Đảm bảo đúng

\(A=1-\frac{2}{3\times5}-\frac{2}{5\times7}-\frac{2}{7\times9}-\cdots-\frac{2}{63\times65}\)

\(=1-\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\cdots+\frac{2}{63\times65}\right)\)

\(=1-\left(\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{63}-\frac{1}{65}\right)\)

\(=1-\left(\frac13-\frac{1}{65}\right)=1-\frac{62}{195}=\frac{133}{195}\)

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

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

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

_Học tốt_

100/101

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

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

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

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

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