10,82896825 :V

a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79

= 1/10 - 1/79

= máy tính ok

mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha

a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)

Bài làm:

a) \(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}\)

\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}\)

\(=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)\)

\(=4\left(\frac{1}{7}-\frac{1}{135}\right)\)

\(=4.\frac{128}{945}=\frac{512}{945}\)

c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}\)

\(=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)\)

\(=4\left(\frac{1}{8}-\frac{1}{506}\right)\)

\(=4.\frac{249}{2024}=\frac{249}{506}\)

d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}\)

\(=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)\)

\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{394}\right)\)

\(=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)

e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}\)

\(=\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)\)

\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)\)

\(=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)

g) Phải sửa \(\frac{1}{802}\)  thành \(\frac{1}{820}\) nhé

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)

\(=1+\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{1}{41.20}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}\right)\)

\(=2\left(1-\frac{1}{41}\right)\)

\(=2.\frac{40}{41}=\frac{80}{41}\)

mù mắt

là sao bạn NGUYỄN HỮU CHUNG 

\(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+........+\frac{1}{78.79}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+............+\frac{1}{78}-\frac{1}{79}\)

\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)

a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)

\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)

\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)

\(=1+1\)

\(=2\)

b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)

\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)

\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)

\(=3+2+2\)

\(=7\)

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)

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

\(=1-\dfrac{1}{7}\)

\(=\dfrac{6}{7}\)

Dấu này * là dấu nhân

Một năm rồi không có ai trả lời à 

THấy cx thương nhưng mk nhìn cái đề thì dài thật cx khó có ai có thời gian mà giải

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

=  1 + 1 + 8

=  2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)

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

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8

= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8

= \(\dfrac{1}{2}\) + 4

= \(\dfrac{9}{2}\) 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

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

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

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

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

  = (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

= 1 + 1 + 8

= 2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)

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

= \(\dfrac{1}{2}\) x 10

= 5

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

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

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

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

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)