HP
1
K
Khách
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18 tháng 9 2020
\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)
<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)
<=> \(3x-\frac{9}{10}=\frac{21}{10}\)
<=> \(3x=3\)
<=> \(x=1\)
NH
15 tháng 7 2023
`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
NP
1
13 tháng 3 2017
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.......+\frac{15}{74.77}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+.......+\frac{3}{74.77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+.....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=5\left(\frac{7}{77}-\frac{1}{77}\right)\)
\(=5.\frac{6}{77}\)
\(=\frac{30}{77}\)
DL
6 tháng 7 2017
a)\(\frac{36^7}{2^{15}\cdot27^5}=\frac{36^7}{\left(2^3\right)^5\cdot27^5}\)
\(=\frac{36^7}{\left(8\cdot27\right)^5}=\frac{36^7}{216^5}\)
\(=\frac{36^7}{36^5\cdot6^5}=\frac{36^5\cdot36^2}{36^5\cdot6^5}\)
\(=\frac{36^2}{6^5}=\frac{\left(6^2\right)^2}{6^5}=\frac{6^4}{6^5}=\frac{1}{6}\)
\(\)
NA
1
8 tháng 11 2016
Gọi phần (.....) là x, ta có:
\(\frac{8}{15}+\frac{4}{15}+\frac{6}{15}=x\cdot\frac{2}{15}\)
\(\Rightarrow\frac{18}{15}=x\cdot\frac{2}{15}\)
\(\Rightarrow x\cdot\frac{2}{15}=\frac{18}{15}\)
\(\Rightarrow x=\frac{18}{15}:\frac{2}{25}\)
\(\Rightarrow x=9\)
Vậy x=9.
W
2
H
12 tháng 4 2019
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}\)
\(B=3.\left(\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(B=3.\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(B=3.\frac{3}{25}\)
\(\Rightarrow B=\frac{9}{25}\)
T
12 tháng 4 2019
\(B=\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}.\)
\(=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)
\(=3\left(\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}\)
\(=\frac{9}{25}\)
Ta có : \(\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-......-\frac{15}{47.45}\)
\(=\frac{3}{8}-\left(\frac{15}{8.11}+\frac{15}{11.14}+\frac{15}{14.17}+......+\frac{15}{47.50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{11}{47}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\frac{1}{8}+\frac{1}{50}\)
\(=\frac{1}{4}+\frac{1}{50}=\frac{27}{100}\)