cho mình hỏi
5/4.6+5/6.8+5/8.10+.....+5/298.300=?
giải giúp mình nhéo cám ơn :))))
K
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HM
1 tháng 4 2016
=5/2(1/4-1/6+1/6-1/8+...+1/208-1/300)
=5/2(1/4-1/300)
=5/2.37/150=37/60
VM
3
T
3 tháng 4 2019
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\frac{37}{150}\)
\(=\frac{37}{60}\)
KN
3 tháng 4 2019
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\frac{37}{150}\)
= \(\frac{37}{60}\)
22 tháng 1 2017
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
NP
4
2
9 tháng 4 2021
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{198.200}\)
\(=\frac{5}{2}\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{198.200}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{5}{2}\left(\frac{50}{100}-\frac{1}{100}\right)\)
\(=\frac{5}{2}.\frac{49}{100}\)
\(=\frac{49}{40}\)
22 tháng 6 2019
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\cdot\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\right)\)
\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{37}{60}\)
27 tháng 6 2020
b) Ta có: \(S=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\)
\(=\frac{1}{2}-\frac{1}{300}=\frac{149}{300}< \frac{200}{300}=\frac{2}{3}\)
hay \(S< \frac{2}{3}\)(1)
Ta có: \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
nên \(\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(vì mỗi ngoặc trên đều có 100 phân số có tử là 1)
\(\Leftrightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}>\frac{1}{200}\cdot100+\frac{1}{300}\cdot100\)
\(\Leftrightarrow Q>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
mà \(\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
nên \(Q>\frac{2}{3}\)
hay \(\frac{2}{3}< Q\)(2)
Từ (1) và (2) suy ra S<Q
LT
11 tháng 3 2020
\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)
\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)
\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)
\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)
24 tháng 7 2020
\(B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{1}{5}B=\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(B-\frac{1}{5}B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}-\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\)\(\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+\frac{4}{11}-\frac{4}{15}+...+\frac{4}{4n-1}-\frac{4}{4n+3}\)
\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{4n-3}\)
\(\frac{4}{5}B=\frac{16n-24}{12n-9}\)
\(B=\frac{\frac{16n-24}{12n-9}}{\frac{4}{5}}\)
\(B=\frac{20n-30}{12n-9}\)
XO
24 tháng 7 2020
B = \(\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}+\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}.\left(\frac{1}{3}-\frac{1}{4n+3}\right)=\frac{5}{12}-\frac{5}{4\left(4n+3\right)}=\frac{5}{12}-\frac{5}{16n+12}\)
10 tháng 4 2017
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(A=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{5}{2}.\frac{99}{100}=\frac{99}{40}\)