\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}...+\frac{1}{605\cdot608}\)
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NT
5
VA
28 tháng 7 2016
Ta có :
\(A=\frac{1}{8.11}+\frac{1}{11.14}+..+\frac{1}{605.608}\)
\(3A=\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)
\(3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)
\(3A=\frac{1}{8}-\frac{1}{608}\)
\(A=\frac{75}{608}:3\)
\(A=\frac{25}{608}\)
Ủng hộ mk nha !!! ^_^
28 tháng 7 2016
Ta có: \(A=\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{605.608}\)
\(\Rightarrow3A=\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{605.608}\)
\(\Rightarrow3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{605}-\frac{1}{608}\)
\(\Rightarrow3A=\frac{1}{8}-\frac{1}{608}=\frac{75}{608}\)
\(\Rightarrow A=\frac{75}{608}:3=\frac{25}{608}\)
Vậy \(A=\frac{25}{608}\)
Ủng hộ cho mik nha bạn?
TM
2
22 tháng 1 2016
A = \(\frac{1}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+.......+\frac{3}{605.608}\right)\)
= \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{605}-\frac{1}{608}\right)\)
= \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{608}\right)=\frac{1}{3}.\frac{76-1}{608}=\frac{1}{3}.\frac{75}{608}=\frac{25}{608}\)
7 tháng 8 2016
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{97.100}\)
\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{97.100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\frac{49}{100}=\frac{49}{300}\)
7 tháng 8 2016
Ta có: \(S=\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{97.100}.\)
\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{97.100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow S=\frac{49}{100}:3=\frac{49}{300}\)
Vậy \(S=\frac{49}{300}\)
CHÚC BẠN HỌC TỐT
NN
28 tháng 6 2016
\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=\frac{2004}{10045}\)
28 tháng 6 2016
\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{1}{11.14}+...+\frac{3}{2006.2009}\)
\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(F=\frac{1}{5}-\frac{1}{2009}\)
\(F=0\)
R
15 tháng 8 2019
a)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540
<=>1/3(3/5.8+3/8.11+...+3/x(x+3) =101/1540
<=>1/3(1/5-1/8+1/8-1/11+...+1/x-1/x+3=101/1540
<=>1/5-1/x+3=303/1540<=>1/x+3=1/308
<=>x+3=308<=>x=305
Nguồn CHTT, hihi !
NT
6
30 tháng 8 2020
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)
30 tháng 8 2020
Tính
\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{17}{34}-\frac{2}{34}=\frac{15}{34}\)
NV
24 tháng 7 2018
\(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)
\(\Leftrightarrow x+3=103\)
\(\Leftrightarrow x\)\(=103-3\)
\(\Leftrightarrow x\)\(=100\)
Vậy x = 100
~~~~~~~Hok tốt~~~~~~~~
24 tháng 7 2018
ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)
\(\Rightarrow x+3=103\)
\(\Rightarrow x=100\)
nhớ k nha
S
18 tháng 3 2017
a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)
=> x + 1 = 1993
x = 1993 - 1
x = 1992
N
18 tháng 3 2017
a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)
3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)
3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)
3A=\(\frac{1}{5}-\frac{1}{608}\)
3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)
Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)
3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)
3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)
3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)
A=\(\frac{201}{3040}\)