1 phần 5.8+ 1 phần 8.11+ 1 phần 11. 14+...+1 phần 197.200
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NT
1
DD
Đoàn Đức Hà
Giáo viên
9 tháng 7 2021
\(\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+...+\frac{1}{197\times200}\)
\(=\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+\frac{14-11}{11\times14}+...+\frac{200-197}{197\times200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{200}\right)\)
\(=\frac{13}{200}\)
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:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\)
\(=\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+\frac{1.3}{11.14.3}+...+\frac{1.3}{197.200.3}\)
\(=\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{200}\right)=\frac{1}{3}.\frac{39}{200}=\frac{13}{200}\)
OY
20 tháng 7 2021
a) Đặt A= \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{36}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{7}{18}\)
A=\(\dfrac{7}{9}\)
NG
3
20 tháng 8 2020
Ta có: \(\left(4,5-2x\right):\frac{3}{4}=1\frac{1}{3}\)
=> \(\left(4,5-2x\right):\frac{3}{4}=\frac{4}{3}\)
=> \(4,5-2x=1\)
=> \(2x=3,5\Rightarrow x=1,75\)
Vậy x=1,75
NT
1
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\(=\frac{-272}{75}\)
NT
2
J
23 tháng 7 2021
B=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{19}{20}\)
\(=\dfrac{1.2.3....19}{2.3.4.....20}\)
\(=\dfrac{1.2.3....19:\left(2.3.....19\right)}{2.3.4.....20:\left(2.3.4.....19\right)}\)
\(=\dfrac{1}{20}\)
Lời giải:
$A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}$
$3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{197.200}$
$3A=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}$
$=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}$
$=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}$
$A=\frac{13}{200}$
Ta có: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{197\cdot200}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{39}{200}=\dfrac{13}{200}\)