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![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}\)
\(S=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}\)
\(S=5\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{5}{16\cdot21}+\frac{5}{21\cdot26}\right]\)
\(S=5\left[1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{26}\right]\)
\(S=5\left[1-\frac{1}{26}\right]=5\cdot\frac{25}{26}=\frac{125}{26}\)
Bài làm
S = \(\frac{5^2}{1.6}\)+ \(\frac{5^2}{6.11}\)+ \(\frac{5^2}{11.16}\)+ \(\frac{5^2}{16.21}\)+\(\frac{5^2}{21.26}\)
S : 5 = \(\frac{5}{1.6}\)+ \(\frac{5}{6.11}\)+ \(\frac{5}{11.16}\) + \(\frac{5}{16.21}\) + \(\frac{5}{21.26}\)
S : 5 = 1 - \(\frac{1}{6}\)+ \(\frac{1}{6}\)- \(\frac{1}{11}\) + \(\frac{1}{11}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{21}\)+ \(\frac{1}{21}\)- \(\frac{1}{26}\)
S : 5 = 1 - \(\frac{1}{26}\)
S : 5 = \(\frac{25}{26}\)
S = \(\frac{125}{26}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{26}-\frac{1}{31}\)
\(\Rightarrow A=1-\frac{1}{31}\)
\(\Rightarrow A=\frac{30}{31}\)
\(A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\)
\(=1+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{11}+\frac{1}{11}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{31-1}{31}=\frac{30}{31}\)
\(\text{Vậy }A=\frac{30}{31}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
A=\(\frac{20^{10}}{20^{10}-1}+\frac{1}{20^{10}-1}\)
B=\(\frac{20^{10}}{20^{10}-3}-\frac{1}{20^{10}-3}\)
Vì 2010-1>2010-3 =>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)
B=0
\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)
\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3-\frac{3}{7}+\frac{3}{13}-\frac{3}{2018}}{7-\frac{7}{20}+\frac{7}{13}-\frac{7}{2018}}\)
\(=\frac{3\left(1-\frac{1}{20}+\frac{1}{13}-\frac{1}{2018}\right)}{7\left(1-\frac{1}{20}+\frac{1}{13}-\frac{1}{2018}\right)}\)
\(=\frac{3}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=20 mủ 10 - 1 +12/(20 mủ 10 -1)=1+12/20 MỦ 10 -1
B=20 mủ 10 - 3 + 2 /(20 mủ 10 - 3)=1+2/20 mủ 10 - 3
Vì ... bạn tự làm nha.nhớ k đấy
A=\(\frac{20^{10}+1}{20^{10}-1}\)=\(\frac{\left(20^{10}-1\right)+2}{20^{10}-1}\)=\(\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}\)=\(1+\frac{2}{20^{10}-1}\)
B= \(\frac{20^{10}-1}{20^{10}-3}=\frac{\left(20^{10}-3\right)+2}{20^{10}-3}\)=\(\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì 2010-1 > 2010-3
=>\(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
=> A < B
Vậy A < B