bn ơi bài này ko có 1/4 đâu: 

đặt A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{36}+\)\(\frac{1}{45}\)

\(\frac{1}{2}A\)=> \(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)

\(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(\frac{1}{2}A\)= \(\frac{1}{4}\)+  \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}A\)= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{5}{20}+\frac{10}{20}-\frac{2}{20}=\frac{13}{20}\)

=> A = \(\frac{13}{20}:\frac{1}{2}=\frac{13}{10}\)

Chúc bn học tốt !

k cho mk 

nếu là 1/4 thì bài  ni ko tính dc

Bài làm:

Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)

\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)

\(=\frac{1}{2}.\frac{11}{12}\)

\(=\frac{11}{24}\)

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)

\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\times\left(1-\frac{1}{12}\right)\)

\(=2\times\frac{11}{12}\)

\(=\frac{11}{6}\)

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\left(1-\frac{1}{12}\right)=2.\frac{11}{12}=\frac{22}{12}=\frac{11}{6}\)

= 7536573657865734657365873464876

a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

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

= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))

= 1 + 1 

= 2

c; \(\dfrac{1}{3}\) + \(\dfrac{-1}{5}\) - \(\dfrac{-1}{4}\) - \(\dfrac{7}{9}\)- \(\dfrac{-5}{12}\) + \(\dfrac{-1}{45}\) + \(\dfrac{1}{41}\)

  = (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{1}{5}\) + \(\dfrac{7}{9}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{9}{45}+\dfrac{35}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{44}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= 1 - 1 + \(\dfrac{1}{41}\)

= 0 + \(\dfrac{1}{41}\)

= \(\dfrac{1}{41}\)

máy tính dâu sao ko lấy ra

Ta có A:B=1.5.6:1:3:5=2

\(P=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

\(\Rightarrow\frac{1}{2}P=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

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

\(=\frac{4}{10}=\frac{2}{5}\)

\(\Rightarrow P=\frac{2}{5}\div\frac{1}{2}\)

\(=\frac{2}{5}.2=\frac{4}{5}\)
Vậy \(P=\frac{4}{5}\).

How to .... >: I know that !!!

\(P=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)

\(\frac{1}{2}P=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(\frac{1}{2}P=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(\frac{1}{2}P=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}P=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(P=\frac{4}{5}\)

quá tuyệt

ta có

(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)

=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)

19/36+17/90+1/12

=(19/36+1/12)+17/90

=7/12+17/90

=105/180+34/180

=139/180

1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45

=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5

=1/1-1/3+1/3-1/2....+1/9-1/5

=1/1

bạn ơ sao mình tính trong máy tính đc 4/5 mà