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\(10A=7+\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{99}}\)
\(10A-A=\left(7+\frac{7}{10}+...+\frac{7}{10^{99}}\right)-\left(\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\right)\)
\(9A=7-\frac{7}{10^{100}}\)
\(A=\frac{7-\frac{7}{10^{100}}}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(3B=2.\frac{96}{721}\)
\(3B=\frac{192}{721}\)
\(\Rightarrow B=\frac{192}{721}:3\)
\(B=\frac{64}{721}\)
\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
Vậy \(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=\frac{2}{3}.\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\frac{96}{721}\)
\(B=\frac{64}{721}\)
Vậy \(B=\frac{64}{721}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\)
\(10A=7+\frac{7}{10}+...+\frac{7}{10^{99}}\)
\(\Rightarrow10A-A=9A=7-\frac{7}{10^{100}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\)
\(\Rightarrow10A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}\)
\(\Rightarrow10A-A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}-\left(\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\right)\)
\(\Rightarrow9A=7-\dfrac{7}{10^{2011}}\)
\(\Rightarrow A=\dfrac{7}{9}.\left(1-\dfrac{1}{10^{2011}}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Help me !
A=7.(1/10+1/10^2+.....+1/10^100)
10A=7.(10/10+10/10^2+....+10/10^100)
10A=7.(1+1/10+....+1/99) 9A=7.(ans-A) 9A=7.(1-1/100) 9A=7-7/100 A=(7-7100):9