\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)

\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

=1/2*4/55

=2/55

Lời giải:

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}+...+\frac{2}{51.55}\)

\(\Rightarrow 2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(=\frac{15-11}{11.15}+\frac{19-15}{15.19}+\frac{23-19}{19.23}+....+\frac{55-51}{51.55}\)

\(=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

\(\Rightarrow A=\frac{2}{55}\)

=\(1\left(\frac{1}{14.15}+\frac{1}{15.19}+......+\frac{1}{51.55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}\right)+\left(\frac{1}{15}-\frac{1}{19}\right).....+\left(\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}....+\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{55}\right)\)

=\(1.\frac{41}{770}\)

=\(\frac{41}{770}\)

a) \(D=\left(2\dfrac{2}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\right)\div\left(-\dfrac{3}{17}\right)\)

\(D=\dfrac{32}{15}\times\dfrac{9}{17}\times\dfrac{3}{32}\times\dfrac{-17}{3}\)

\(D=\dfrac{-3}{5}\)

b) \(\dfrac{1}{2}-\dfrac{1}{3\times7}-\dfrac{1}{7\times11}-\dfrac{1}{11\times15}-\dfrac{1}{15\times19}-\dfrac{1}{19\times23}-\dfrac{1}{23\times27}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+\dfrac{1}{15\times19}+\dfrac{1}{19\times23}+\dfrac{1}{23\times25}\right)\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+\dfrac{4}{11\times15}+\dfrac{4}{15\times19}+\dfrac{4}{19\times23}+\dfrac{4}{23\times27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\left[\dfrac{1}{4}\left(\dfrac{9-1}{27}\right)\right]\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\times\dfrac{8}{27}\)

\(=\dfrac{1}{2}-\dfrac{2}{27}\)

\(=.....\)

Đó đến đây bn tự lm nốt. Câu c bn lm tương tự.

Mình cho bn dạng này, nếu bn chưa biết (để lm câu c)

\(\dfrac{x}{y\left(y+x\right)}=\dfrac{x}{y}-\dfrac{x}{y+x}\)

Chúc bn học tốt

bạn tham khảo nha, cách làm như vậy đó

Câu hỏi của Nguyễn Thị Mai Ca - Toán lớp 7 - Học toán với OnlineMath 

ban kia lam dung roi do

k tui nha 

thanks

\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)

= \(4.\left(\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)

=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)

= \(1.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)

=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)

= \(1.\dfrac{8}{27}\)

= \(\dfrac{8}{27}\)

\(S=\dfrac{1}{2}-\dfrac{1}{3.7}-\dfrac{1}{7.11}-...........-\dfrac{1}{23.27}\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+..........+\dfrac{1}{23.27}\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+.......+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

\(=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\)

\(=\dfrac{1}{2}-\dfrac{8}{27}\)

\(=\dfrac{11}{54}\)

Bạn xem lại đề bài đi chứ thế này thì cần j phải so sánh nx

Này nhé: đã có \(\dfrac{1}{2}=2^{-1}\) mà \(2^{-1}< 2^{51}\) là điều quá rõ rồi

Đã thế lại còn trừ liên hoàn từ... (đấy nói chung là phần sau) thì rõ ràng hiển nhiên là \(S< 2^{51}\) còn cái j nx

Chúc bn học tốt

Bạn tham khảo bài làm của mình nhé !!

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}...+\frac{2}{51.55}\)

\(\Leftrightarrow2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{4}{55}\)

\(\Leftrightarrow A=\frac{4}{110}\)