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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-...-\frac{1}{110}\)
\(=\)\(1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=\)\(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=\)\(1-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=\)\(1-\left(1-\frac{1}{11}\right)\)
\(=\)\(1-1+\frac{1}{11}\)
\(=\)\(\frac{1}{11}\)
Chúc bạn học tốt ~
\(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{110}\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=1-\left(1-\frac{1}{11}\right)\)
\(=1-\frac{10}{11}\)
\(=\frac{1}{11}\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Đặt\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}...+\frac{1}{110}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{10.11}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
\(S=1-\frac{1}{11}\)
\(S=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ... + \(\frac{1}{10.11}\)
= \(\frac{1}{1}\)-\(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+... + \(\frac{1}{10}\)-\(\frac{1}{11}\)
= \(\frac{1}{1}\)-\(\frac{1}{11}\)
= \(\frac{10}{11}\)
ai tốt bụng thì tk mk nha, mk đg âm điểm đây
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\)\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{4}\)-\(\frac{1}{4}\)+\(\frac{1}{5}\)-....+\(\frac{1}{10}\)=x-\(\frac{113}{260}\)
\(\Leftrightarrow\)x-\(\frac{113}{260}\)=\(\frac{1}{10}\)
\(\Leftrightarrow\)x=\(\frac{139}{260}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vế trái lớn hơn hoặc bằng 0 nên 11x lớn hơn hoặc bằng 0.
\(\Rightarrow x\ge0\)
Do vậy chỉ cần bỏ dấu giá trị tuyệt đối là tính được.
Kết quả cuối cùng được \(x=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/10 - 1/11
= 1 - 1/11
= 10/11
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
Tham khảo nhé~