1 phần 20 + 1 phần 30 + 1phaanf 42 ... + 1 phần 90 = ???
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\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+....+1-\dfrac{1}{90}=1+1+...+1-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=9-\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{9x10}\right)=9-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)=9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
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\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
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\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}\right)\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8-1}{8}=\dfrac{7}{8}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
`=`\(1-\dfrac{1}{8}\)
`=`\(\dfrac{7}{8}\)
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\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(B=\dfrac{1}{2}-\dfrac{1}{9}\)
\(B=\dfrac{9}{18}-\dfrac{2}{18}\)
\(B=\dfrac{7}{18}\)
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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)
A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)+ \(\dfrac{1}{29\times30}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)+ \(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)
A = 1 - \(\dfrac{1}{30}\)
A = \(\dfrac{29}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(S=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(S=\frac{1}{3}-\frac{1}{8}\)
\(S=\frac{8}{24}-\frac{3}{24}\)
\(S=\frac{5}{24}\)
\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(S=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(S=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
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\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{90}\)
\(=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{5-2}{20}=\dfrac{3}{20}\)
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