Sửa đề :\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)

\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{15}\right)\)

\(=\frac{5}{11}-\frac{1}{3}\)

\(=\frac{15-11}{33}=\frac{14}{33}\)

\(A = {5\over10}-{5\over11}+{5\over11}-{5\over12}+...+{5\over14}-{5\over15} = {5\over10}-{5\over15}={1\over6} \)Ban chep de thieu nha^^

\(\frac{5}{10.11}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)

\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)

\(=5\left(\frac{1}{10}-\frac{1}{15}\right)\)

\(=5.\frac{5}{150}\)

\(=\frac{25}{150}=\frac{1}{6}\)

A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15

A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5

A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5

A= (1/10-1/15):5

A= 1/30:5

A= 1/150

cậu kia làm sai rùi

A .1/5 = 1/10.11+1/11.12+1/12.13+...+1/99,100

A . 1/5 = 1/10-1/11+1/11-1/12+...+1/99-1/100

A .1/5 = 1/10-1/100

A.1/5 = 9/199

A = 9/20

k nhé

\(A=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+.....+\frac{5}{99.100}\)

=\(\frac{5}{10}-\frac{5}{11}+\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}\)

=\(\frac{5}{10}-\frac{5}{100}=\frac{45}{100}\)=\(\frac{9}{20}\)

\(A=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+.....+\frac{5}{99.100}\)

     \(=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{99.100}\right)\)

    \(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\right)\)

   \(=5.\left(1-\frac{1}{100}\right)\)

   \(=5.\frac{99}{100}=\frac{99}{20}\)

b)

\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)

\(5\left(1-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{31}\right)\)

\(5\left(1-\frac{1}{31}\right)\)

\(=\frac{150}{31}\)

a)

\(\frac{7}{10}-\frac{7}{11}+...+\frac{7}{69}-\frac{7}{70}\)

\(=\frac{7}{10}-\frac{7}{70}\)

\(=\frac{3}{5}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(C=7.\frac{3}{35}\)

\(C=\frac{3}{5}\)

\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)

= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)

= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)

\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)

\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)

\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)

\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)

\(=\dfrac{3}{1}.\dfrac{1}{10}\)

\(=\dfrac{3}{10}\)

A=.....

=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY

TỰ LÀM NHE

\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)

\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)

\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)

\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)

\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)

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

\(C=9-\left(1-\frac{1}{10}\right)\)

\(C=9-\frac{9}{10}=\frac{81}{10}\)

các bạn đều sai hết mà các bạn lại cho những câu hỏi đó đúng 

vãi nồi :

S=7/10.11+7/11.12+...+7/69.70

=>S= 7.(1/10.11+1/11.12+...+1/69.70)

=> S=  7. ( 1/10-1/11+...+1/69-1/70)

=> S= 7. ( 1/10 - 1/70)

=> S = 7. 3/35

=> S = 3/5

We are one_Nguyễn Ngọc Sáng

\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}=\frac{6}{10}=\frac{3}{5}\)

= 7.( \(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-....-\frac{1}{70}\))

= 7.( \(\frac{1}{10}-\frac{1}{70}\))

= 7.(\(\frac{7}{70}-\frac{1}{70}\))

= 7.\(\frac{6}{70}\)

= \(\frac{3}{5}\)