so sánh A với 1 , biếtA = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
![](https://rs.olm.vn/images/avt/0.png?1311)
- A ở trên giữa các phân số là dấu " + " nha mấy bạn !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)
\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)
\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)
\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)
=> \(x=2\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)
\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)
\(=>x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(A=\frac{1}{2}-\frac{1}{17}\)
\(A=\frac{15}{34}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
B=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}+\frac{3}{17.20}\)
\(\Rightarrow B=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{20}=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=3.\left(1-\frac{1}{10}\right)\)
\(A=3.\frac{9}{10}=\frac{27}{10}\)
\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)
\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)
\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)
Vậy:..........................................(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{x}{2}-\frac{x}{5}+\frac{x}{5}-\frac{x}{8}+\frac{x}{8}-\frac{x}{11}+\frac{x}{11}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{x}{2}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{7x}{14}-\frac{x}{14}=\frac{1}{21}\)
\(\Leftrightarrow\frac{6x}{14}=\frac{1}{21}\)
\(\Leftrightarrow126x=14\)
\(\Leftrightarrow x=\frac{1}{9}\)
Học tôt
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
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