mù mắt

là sao bạn NGUYỄN HỮU CHUNG 

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

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

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

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

=1/16

MÌNH KO CHẮC NHÉ

(1/10+-1/10)+(1/11+-1/11)+(1/12+-1/12)+(-1/13+1/13)+(-1/14+1/14)+(-1/15+1/15)+1/16

=0 + 0 +0 + 0 +0 +0 +1/16

=1/16

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

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

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

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

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

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

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

Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

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

\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)

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

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

\(=\dfrac{1}{9}-\dfrac{1}{15}\)

\(=\dfrac{2}{45}\)

-Chúc bạn học tốt-

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

= \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\)

= \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

= \(\dfrac{1}{9}-\dfrac{1}{15}\)

= \(\dfrac{2}{45}\)

\(S=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{99.100}\)

\(\Rightarrow S=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow S=\frac{1}{10}-\frac{1}{100}\)

\(\Rightarrow S=\frac{99}{100}\)

\(S=\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{99.100}\)

\(=\frac{11-10}{10.11}+\frac{12-11}{11.12}+...+\frac{100-99}{99.100}\)

\(=\frac{11}{10.11}-\frac{10}{10.11}+\frac{12}{11.12}-\frac{11}{11.12}+....+\frac{100}{99.100}-\frac{99}{99.100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}=\frac{9}{100}\)

Với n > 0 Ta có:

\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)

\(=\sqrt{n+1}+\sqrt{n}\)

\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)

\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)

\(\sqrt{16}+\sqrt{9}=3+4=7\)