tính b=3/10+3/40+3/88+...+3/340
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28 tháng 4 2016
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!
28 tháng 4 2016
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 \(A=\frac{27}{20}\)
HT
0
JK
27 tháng 7 2019
\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)
\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)
\(=\frac{1}{2}-\frac{1}{23}\)
\(=\frac{21}{46}\)
A
1
17 tháng 5 2020
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 : \(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{1}{340}\)
\(C=3\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}\right)+\frac{1}{340}\)
\(C=3.\frac{3}{22}+\frac{1}{340}=\frac{9}{22}+\frac{1}{340}=\frac{1541}{3740}\)
18 tháng 3 2018
S = 1/2-1/5+1/5-1/8+1/8-...-1/20
S = 1/2-1/20
S = 9/20 nha
LK
18 tháng 3 2018
\(S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
ST
2
3 tháng 4 2016
\(A=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
3 tháng 4 2016
\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)
= \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{20}\right)\)
= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
= 1. \(\frac{9}{20}\)
= \(\frac{9}{20}\)
N
2
16 tháng 5 2018
\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3A=3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\left(\frac{1}{2}-\frac{1}{20}\right)\div3=\frac{9}{20}\div3=\frac{9}{20.3}=\frac{3}{20}\)
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......
16 tháng 5 2018
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
JK
27 tháng 7 2019
\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)
\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)
\(=\frac{1}{2}-\frac{1}{23}\)
\(=\frac{21}{46}\)
KQ là 9/20 bạn nha
Chúc bạn học tốt
b = 3/2.5+3/5.8+3/8.11+.....+3/17.20
= 1/2.5+1/5.8+.......+1/17.20
= 1/2-1/5+1/5-1/8+......+1/17-1/20
= 1/2- 1/20
= 9/20