\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}\right)\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8-1}{8}=\dfrac{7}{8}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

`=`\(1-\dfrac{1}{8}\)

`=`\(\dfrac{7}{8}\)

sos 

cần gấp nha mn

\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)

\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)

\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(B=\dfrac{1}{2}-\dfrac{1}{9}\)

\(B=\dfrac{9}{18}-\dfrac{2}{18}\)

\(B=\dfrac{7}{18}\)

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 = \(\dfrac{29}{30}\)

29/30 nha

81

\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+....+1-\dfrac{1}{90}=1+1+...+1-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=9-\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{9x10}\right)=9-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)=9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)

đề thiếu bn ơi

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)\(+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)\(+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72

=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

=1-1/9

=8/9

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7

=1-1/7

=6/7

\(\frac{113}{140}\frac{ }{ }\)

=\(\frac{6}{7}\)

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)

CHÚC BN HỌC TỐT !!! ^_^

1)

A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

   = \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)

   = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

   = \(\frac{1}{5}-\frac{1}{12}\)

   = \(\frac{7}{60}\)

B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

   = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

   = \(\frac{3.4.5.....100}{2.3.4....99}\)

   = \(\frac{100}{2}=50\)

C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)

   = \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)

   = \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)

   = \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)

   = \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

~ Hok tốt ~

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

....