= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)

 = \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

 = \(\dfrac{5}{2}.\dfrac{100}{101}\)

 = \(\dfrac{250}{101}\)

1) âm năm phần 12

2) âm mười bảy phần 9

3) -1 

Đây là đáp án còn làm bài từ làm nhé

a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)

\(=\dfrac{3}{5}+\dfrac{41}{25}\)

\(=\dfrac{15}{25}+\dfrac{41}{25}\)

\(=\dfrac{56}{25}\)

a) 

b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

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

\(=\) \(1-\dfrac{1}{10}\)

\(=\dfrac{-9}{10}\)

B= \(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}\)

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

= 1+\(\dfrac{1}{8}\)=\(\dfrac{9}{8}\)

A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)

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

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

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

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

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

\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)

8/9

a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

\(a,\dfrac{-8}{15}+\dfrac{13}{30}-\dfrac{5}{12}=\dfrac{-32}{60}+\dfrac{26}{60}-\dfrac{25}{60}=-\dfrac{31}{60}\\ b,\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}:\dfrac{11}{30}\right)=\dfrac{21}{4}+\left(\dfrac{-5}{6}+\dfrac{3}{11}\right)=\dfrac{21}{4}+\dfrac{-37}{66}=\dfrac{619}{132}\)

\(c,\dfrac{-20}{21}.\dfrac{22}{35}+\dfrac{-20}{21}.\dfrac{13}{35}+\dfrac{-22}{21}=\dfrac{-20}{21}\left(\dfrac{22}{35}+\dfrac{13}{35}\right)+\dfrac{-22}{21}=\dfrac{-20}{21}.1+\dfrac{-22}{21}=\dfrac{-20}{21}+\dfrac{-22}{21}=\dfrac{-42}{21}=-2\)

1

a)103/35

b)-3/13