-Ta có: \(\dfrac{1}{2}=\dfrac{10}{20}=\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có 10 số \(\dfrac{1}{20}\)).

Mà \(\dfrac{1}{20}< \dfrac{1}{19};\dfrac{1}{20}< \dfrac{1}{18};...;\dfrac{1}{20}< \dfrac{1}{11}\)

\(\Rightarrow\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}< \dfrac{1}{20}+\dfrac{1}{19}+\dfrac{1}{18}+...+\dfrac{1}{11}\)

\(\Rightarrow\dfrac{1}{2}< S\)

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

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

Dấu này * là dấu nhân

Một năm rồi không có ai trả lời à 

\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)

\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)

\(A\)\(=\)\(\frac{2}{45}\)

\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)

Sau đó nhân phân phối ra là xong nhé bạn 

a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)

b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)

c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)

d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)

\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)

\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)

\(=\dfrac{25}{14}\)

\(b.-\dfrac{7}{14}:\left(-0,14\right)\)

\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)

\(=\dfrac{25}{7}\)

\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)

\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)

\(=-\dfrac{2}{3}\)

\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)

\(=\dfrac{15}{7}:\dfrac{15}{14}\)

\(=2\)

mù mắt

là sao bạn NGUYỄN HỮU CHUNG 

Để mình giải cho!

Số số hạng của tổng trên là: 

\(\left(2020-11\right):1+1=2010\)

Tổng trên là: 

\(\left(2020+11\right).2010:2=2041155\)

                Bài giải:

Số số hạng của dãy trên là:

   \(\left(2020-11\right):1+1=2010\)

Tổng dãy số trên là:

   \(\left(2020+11\right).2010:2=2041155\)

Đáp số: \(2041155\)

mình chỉ tính vế bên thôi nhé bn tích x

\(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{25.28}\)

\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.....+\frac{1}{25}-\frac{1}{28}\right)\)

=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)

=\(???\)

ban oi sai roi

a) B = \(\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{21}{20}=\dfrac{1}{3}.1.....\dfrac{21}{1}=\dfrac{21}{3}=7\)

b) Em chịu, chưa học số âm :)