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X
6
27 tháng 8 2019
\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)
\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\cdot\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
XO
27 tháng 8 2019
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
HN
4
OY
5 tháng 9 2021
Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)
\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)
5 tháng 9 2021
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(=\dfrac{1}{2}-\dfrac{1}{98}\)
\(=\dfrac{24}{49}\)
QH
1
PT
1
15 tháng 7 2018
Ta có : A = 1/ 2.5 + 1/ 5.8 + 1/ 8.11 + ... + 1/ (3n-1).(3n+2) .
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/ 3n-1 - 1/ 3n+2 .
= 1/2 - 1/ 3n+2 .
= 3n + 2 - 2 / 2 .( 3n+2 ) .
= 3n / 2.(3n+2) .
KH
2
25 tháng 9 2016
<=> \(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=>\(x+3=308\)
<=>\(x=305\)
AP
3
21 tháng 8 2017
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{90.93}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{90.93}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{90}-\dfrac{1}{93}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{93}\right)\)
\(=\dfrac{91}{558}\)
NQ
1
19 tháng 1 2018
Đặt A=1/2.5+1/5.8+...+1/(3n-1)(3n+2)
3A=3/2.5+3/5.8+....+3/(3n-1)(3n+2)
3A=1/2-1/5+1/5-1/8+....+1/3n-1-1/3n+2
3A=1/2-1/3n+2
3A=3n/6n+4
A=(3n/6n+4) /3
A=n/6n+4(đpcm)
TK
1
5 tháng 8 2020
b/ Ta có :
\(M=\frac{3^2}{2.5}+\frac{3^2}{5.8}+....+\frac{3^2}{98.101}\)
\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{98.101}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{98}-\frac{1}{101}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=3.\frac{99}{202}\)
\(=\frac{297}{202}\)
Vậy....
1/2.5 nhé
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)
\(3A=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{101-98}{98.101}\)
\(3A=\frac{5}{2.5}-\frac{2}{2.5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)
\(3A=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
\(\Leftrightarrow A=\frac{99}{202}\div3\)
\(\Rightarrow A=\frac{33}{202}\)