\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}......1\frac{1}{120}\) = ?
K
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T
16 tháng 2 2017
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{120}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{11.11}{10.12}\)
\(=\frac{2.2.3.3.4.4...11.11}{1.3.2.4.3.5...10.12}\)
\(=\frac{\left(2.3.4...11\right)\left(2.3.4...11\right)}{\left(1.2.3...10\right).\left(3.2.5...12\right)}\)
\(=\frac{11.2}{1.12}\)
\(=\frac{11}{6}\)
NT
0
2 tháng 9 2016
\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...1\frac{1}{120}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{11.11}{10.12}\)
\(=\frac{2.3.4...11}{1.2.3...10}.\frac{2.3.4...11}{3.4.5...12}\)
\(=11.\frac{2}{12}=11.\frac{1}{6}=\frac{11}{6}\)
HN
7 tháng 3 2017
\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{121}{120}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.\dfrac{11.11}{10.12}\)
\(=\dfrac{2}{1}.\dfrac{11}{12}=\dfrac{11}{6}\)
NK
Tìm x biết:
\(\frac{x}{2013}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-.....-\frac{1}{120}=\frac{5}{8}\)
0
NT
4
4 tháng 3 2017
Ta có:
\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1
=> \(x=2013\)
Vậy \(x=2013\)
VM
12 tháng 11 2019
Ta có:
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=1.2008\)
\(\Rightarrow x=2008\)
Vậy \(x=2008.\)
Chúc bạn học tốt!
25 tháng 6 2016
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-\left(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)
\(\frac{x}{2008}-2.\frac{3}{16}=\frac{5}{8}\)
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\frac{x}{2008}=\frac{5}{8}+\frac{3}{8}\)
\(\frac{x}{2008}=1=\frac{2008}{2008}\)
=> x = 2008
Vậy x = 2008