\(\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+...+\dfrac{5}{110}\)

\(=\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+...+\dfrac{5}{10.11}\)

\(=\dfrac{5}{1}-\dfrac{5}{2}+\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{5}{3}-\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{5}{5}+...+\dfrac{5}{10}-\dfrac{5}{11}\)

\(=\dfrac{5}{1}-\dfrac{5}{11}=\dfrac{50}{11}\)

=5(1/2+1/6+1/12+1/20+1/30+1/42+...+1/110)

=5(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/10.11)

=5(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/10-1/11)

=5(1-1/11)=5.10/11=50/11

\(\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+....+\dfrac{5}{110}\)

\(=\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+....+\dfrac{5}{10.11}\)

\(=\dfrac{5}{1}-\dfrac{5}{2}+\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{5}{3}-\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{5}{5}+....+\dfrac{5}{10}-\dfrac{5}{11}\)

\(=\dfrac{5}{1}-\dfrac{5}{11}\)

\(=\dfrac{50}{11}\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{110}\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\dfrac{10}{11}=\dfrac{100}{11}\)

=>-2x=1967/220

hay x=-1967/440

bn đổi thành (1-1/6)+(1-1/12)+.....+(1-1/110)

=9-(1/2.3+1/3.4+...+1/10.11)

đến đây bn tự lằm nha

a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)

b) Bạn làm tương tự. 

mnbvcxzasdf

976554

1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 .....+1 - 1/110= 10 - ( 1/2 + 1/6 + ...+ 1/110) = 10 - ( 1 - 1/ 2+ 1/2 - 1/ 3+ 1/3 - 1/4 ....+ 1/10 - 1/11)= 10 - (1 - 1/11)= 10 - 10/11 = 100/11 

= 100/11 nha bạn

k cho minh nha

chuc ban hoc tot

= \(\frac{100}{11}\)

k cho mik nha 

thank you

a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)

Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1

\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=8-\frac{2}{5}=\frac{38}{5}\)

b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)

Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1

\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{11}\right)\)

\(=10-\frac{10}{11}=\frac{100}{11}\)

100/11 nha

câu b bằng 891/110

câu a bằng 76/10