Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(=>A=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{17\cdot20}\)

\(=>A=\frac{9}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)

\(=>A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(=>A=3\left(\frac{1}{2}-\frac{1}{20}\right)=3\left(\frac{10}{20}-\frac{1}{20}\right)=3\cdot\frac{9}{20}=\frac{27}{20}\)

Chúc bạn học tốt!

Chọn mình nhé !  

Ta có:

\(A=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(\Rightarrow A=3\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\right)\)

\(\Leftrightarrow A=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(\Leftrightarrow A=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(\Leftrightarrow A=3\left(\frac{1}{2}-\frac{1}{20}\right)=3.\frac{9}{20}=\frac{27}{20}\)

Vậy \(A=\frac{27}{20}\)

để mình giúp

\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)

\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)

\(=\frac{1}{2}-\frac{1}{23}\)

\(=\frac{21}{46}\)

Bài này có 2 cách, nhưng cách nào thì cách cx phải dùng tới máy tính, ai cs cách hay show hộ kham khảo ! 

Cách 1 : cầm máy tính lên bấm 

Cách 2 : \(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{1}{340}\)

\(C=3\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}\right)+\frac{1}{340}\)

\(C=3.\frac{3}{22}+\frac{1}{340}=\frac{9}{22}+\frac{1}{340}=\frac{1541}{3740}\)

S = 1/2-1/5+1/5-1/8+1/8-...-1/20

S = 1/2-1/20

S = 9/20 nha

\(S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

co ai ra e qua  khac ko?

\(A=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)

\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

= \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{20}\right)\)

= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

= 1. \(\frac{9}{20}\)

= \(\frac{9}{20}\)

\(=\frac{27}{20}\)nhé!

^_^

kb đi kb đi kb đi NHA

\(\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(=3.\left(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\right)\)

\(=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(=3.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=3.\frac{9}{20}=\frac{27}{20}\)

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)

\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(3A=\frac{1}{2}-\frac{1}{20}\)

\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)

Vậy ................

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{9999}{10000}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{99.101}{100.100}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right).\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right).\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)

\(B=\frac{1\cdot2\cdot3\cdot..\cdot99}{2\cdot3\cdot4\cdot..\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)

\(B=\frac{1}{100}\cdot\frac{101}{2}=\frac{101}{200}\)

vậy......

A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20

A=1/3.(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17+3/17.20)

A=1/3.(1/2-1/20)

=3/20

B=1.3/2.2+2.4/3.3+3.5/4.4+...+99.101/100.100

B=(1.2.3...99).(3.4.5...101)/(2.3.4...100).(2.3.4...100)

B=\(\frac{1.2....99}{2.3...100}\).\(\frac{3.4...101}{2.3...100}\)

B=1/100.101/2=101/200

\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)

\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)

\(=\frac{1}{2}-\frac{1}{23}\)

\(=\frac{21}{46}\)