NT
1
K
Khách
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AH
Akai Haruma
Giáo viên
9 tháng 7 2021
Lời giải:
$A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}$
$3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{197.200}$
$3A=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}$
$=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}$
$=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}$
$A=\frac{13}{200}$
9 tháng 7 2021
Ta có: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{197\cdot200}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{39}{200}=\dfrac{13}{200}\)
NT
2
AD
9 tháng 7 2021
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{197}-\frac{1}{200}\)
\(=\frac{1}{5}-\frac{1}{200}\)
\(=\frac{40}{200}-\frac{1}{200}\)
\(=\frac{39}{200}\)
#H
QA
10 tháng 7 2021
Trả lời:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\)
\(=\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+\frac{1.3}{11.14.3}+...+\frac{1.3}{197.200.3}\)
\(=\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{200}\right)=\frac{1}{3}.\frac{39}{200}=\frac{13}{200}\)
8 tháng 2 2019
a) = ( 9/10 - 4/5) + ( 3/4 + 1/2)
= ( 9/10 - 8/10) + ( 3/4 + 2/4)
= 1/10 + 5/4
= 2/20 + 25/20 = 27/20
b) = (4/5 + 4/15) + (5/6 - 1/2)
= (12/15 - 4/15) + (5/6 - 3/6)
= 8/15 + 2/6
= 16/30 + 10/30
= 26/30 = 13/15
c) = 55 - 39 - 1 + 60 /75 = 75/75 = 1
d) = 47 + 35 - 36 + 15/42 = 61/42
BM
1
VD
3 tháng 4 2018
x.\(\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)=-1\frac{3}{5}\)
x.\(\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{1}{2}-\frac{1}{17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{17}{34}-\frac{2}{34}\right)=\frac{-8}{5}\)
x.\(\frac{15}{34}=\frac{-8}{5}\)
x\(=\frac{-8}{5}:\frac{15}{34}\)
x\(=\frac{-8}{5}.\frac{34}{15}\)
x\(=\frac{-272}{75}\)
Vậy x\(=\frac{-272}{75}\)
LT
6
55
17 tháng 6 2019
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
\(=\frac{1}{5}-\frac{1}{108}\)
\(=\frac{108}{540}-\frac{5}{540}=\frac{103}{540}\)
17 tháng 6 2019
Ta có:
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)
= \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)
= \(\frac{1}{5}-\frac{1}{108}\)
= \(\frac{103}{540}\)
NT
3
\(\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+...+\frac{1}{197\times200}\)
\(=\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+\frac{14-11}{11\times14}+...+\frac{200-197}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=\frac{13}{200}\)