Em cần phần nào nhỉ .

A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)

A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)

A = \(\dfrac{105}{106}\)

B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)

B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)

C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))

C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)

C = \(\dfrac{5}{51}\) 

D = \(\dfrac{1}{2}\) +   \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)

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

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)

D = \(\dfrac{8}{9}\)

E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))

E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)

E = \(\dfrac{147}{200}\)

Giúp với 

Ta có:\(\dfrac{20}{2\times7}+\dfrac{20}{7\times12}+\dfrac{20}{12\times17}+\dfrac{20}{17\times22}+\dfrac{20}{22\times27}+\dfrac{20}{27\times32}\)

 \(=4\times\left(\dfrac{5}{2\times7}+\dfrac{5}{7\times12}+\dfrac{5}{12\times17}+\dfrac{5}{17\times22}+\dfrac{5}{22\times27}+\dfrac{5}{27\times32}\right)\)

    \(=4\times\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{32}\right)\)

\(=4\times\left(\dfrac{1}{2}-\dfrac{1}{32}\right)=4\times\dfrac{15}{32}=\dfrac{30}{16}\)

\(=5+\dfrac{4}{5}\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+...+\dfrac{5}{52\cdot57}\right)\)

\(=5+\dfrac{4}{5}\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{52}-\dfrac{1}{57}\right)\)

\(=5+\dfrac{4}{5}\cdot\dfrac{55}{114}=\dfrac{307}{57}\)

\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{102-97}{97.102}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{97}-\dfrac{1}{102}\)

\(=\dfrac{1}{2}-\dfrac{1}{102}=\dfrac{25}{51}\)

\(=\dfrac{7-2}{2.7}+\dfrac{12-7}{7.12}+...+\dfrac{97-92}{92.97}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}\)

\(=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)

a, 3/5 x 7/9 + 11/9 x 3/5 = 3/5 x (7/9 + 11/9) = 3/5 x 2 = 6/5

b, 14/13 : 1/2 + 3/13 x 2 - 2 x 7/13

= 14/13 x 2 + 3/13 x 2 - 2 x 7/13

= 2 x ( 14/13 + 3/13 - 7/13)

= 2 x 10/13 = 20/13

14/13:1/2+3/13x2-2x7/13=(14/13+3/13-7/13)x2=10/13x2=20/13

riêng 3/1*2 là ra 6 rồi thì đương nhiên C>2

\(C=\dfrac{10}{7\cdot12}+\dfrac{10}{12\cdot17}+...+\dfrac{10}{502\cdot507}\)

\(=2\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)

\(=2\cdot\dfrac{500}{3549}=\dfrac{1000}{3549}\)