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

=> \(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)    (dấu . có nghĩa là nhân)

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

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

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

\(=\frac{1}{2}-\frac{1}{20}\) 

\(=\frac{9}{20}\)            

Đây là kiến thức lớp 6 nhá =)) bạn mà có chỗ nào ko hiểu thì hỏi ng thầy cô giạy bạn ý

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

\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)

\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+\frac{3}{17\cdot20}\right)\)

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

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

\(=\frac{1}{3}\cdot\frac{9}{20}\)

\(=\frac{3}{20}\)

A =  1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340 

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

3 A = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/17.20

3A = 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20

3A = 1/2 - 1/20

3A = 9/20

A = 9/20 : 2

1/10+1/40+1/88+1/154+1/238+1/340

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

=1/2-1/5+1/5-1/8+1/8-...+1/17-1/20

=1/2-1/20

=9/20

​k cho mình nha thanks bạn

= 3/30 nhé bạn

\(\frac{3}{20}\)

k mk nhé 

bạn có thể trả lời đầy đủ ko

8,112

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

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

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

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

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

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

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

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

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

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

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

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

Ta có

Thằng ngủ ko biết làm bài dễ vãi

Đầu buồi 

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

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{20}\right)\)

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