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Ta có : \(\frac{15}{5.8}-\frac{15}{8.11}-\frac{15}{11.14}-......-\frac{15}{47.45}\)
\(=\frac{3}{8}-\left(\frac{15}{8.11}+\frac{15}{11.14}+\frac{15}{14.17}+......+\frac{15}{47.50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+.....+\frac{11}{47}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\left(\frac{1}{8}-\frac{1}{50}\right)\)
\(=\frac{3}{8}-\frac{1}{8}+\frac{1}{50}\)
\(=\frac{1}{4}+\frac{1}{50}=\frac{27}{100}\)
`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`
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.......+\frac{15}{74.77}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+.......+\frac{3}{74.77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+.....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=5\left(\frac{7}{77}-\frac{1}{77}\right)\)
\(=5.\frac{6}{77}\)
\(=\frac{30}{77}\)
a)\(\frac{36^7}{2^{15}\cdot27^5}=\frac{36^7}{\left(2^3\right)^5\cdot27^5}\)
\(=\frac{36^7}{\left(8\cdot27\right)^5}=\frac{36^7}{216^5}\)
\(=\frac{36^7}{36^5\cdot6^5}=\frac{36^5\cdot36^2}{36^5\cdot6^5}\)
\(=\frac{36^2}{6^5}=\frac{\left(6^2\right)^2}{6^5}=\frac{6^4}{6^5}=\frac{1}{6}\)
\(\)
a) \(\frac{7}{15}+\frac{9}{10}+\frac{8}{15}-\frac{-1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\frac{7}{15}+\frac{9}{10}+\frac{8}{15}+\frac{1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{9}{10}+\frac{1}{10}\right)-2+\frac{1}{157}\)
\(=1+1-2+\frac{1}{157}\)
\(=2-2+\frac{1}{157}\)
\(=0+\frac{1}{157}=\frac{1}{157}\)
b) \(\frac{1}{13}+\frac{16}{7}+\frac{3}{105}-\frac{9}{7}-\frac{-12}{13}\)
\(=\frac{1}{13}+\frac{16}{7}+\frac{1}{35}-\frac{9}{7}+\frac{12}{13}\)
\(=\left(\frac{1}{13}+\frac{12}{13}\right)+\left(\frac{16}{7}-\frac{9}{7}\right)+\frac{1}{35}\)
\(=1+1+\frac{1}{35}\)
\(=2+\frac{1}{35}\)
\(=\frac{70}{35}+\frac{1}{35}=\frac{71}{35}\)
\(x-\frac{1}{15}=\frac{1}{10}\)
\(x=\frac{1}{10}+\frac{1}{15}\)
\(x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
\(-\frac{2}{15}-x=-\frac{3}{10}\)
\(x=-\frac{2}{15}+\frac{3}{10}\)
\(x=\frac{1}{15}\)
Vậy \(x=\frac{1}{15}\)
\(x-\frac{1}{15}=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+\frac{1}{15}\)
\(\Rightarrow x=\frac{1}{6}\)
\(\frac{-2}{15}-x=\frac{-3}{10}\)
\(\Rightarrow x=\frac{-2}{15}-\frac{-3}{10}\)
\(\Rightarrow x=\frac{1}{6}\)
\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)
<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)
<=> \(3x-\frac{9}{10}=\frac{21}{10}\)
<=> \(3x=3\)
<=> \(x=1\)