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1/10 A =7/10^2+7/10^3+..............+7/10^2020
9/10*A=(7/10+7/10^2+......................+7/10^2019)-(7/10^2+7/10^3+........+7/10^2020)
=7/10-7/10^2020
A=10/9 .(7/10-7/10^2020)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\)
\(10A=7+\frac{7}{10}+...+\frac{7}{10^{99}}\)
\(\Rightarrow10A-A=9A=7-\frac{7}{10^{100}}\)
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a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
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D=777777777777777777777777777777777777777777777.../1000000000000000000000000000000000000000000...
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Câu 1:
\(S=\frac{10}{7}+\frac{10}{7^2}+\frac{10}{7^3}+...+\frac{10}{7^{10}}\)
\(\frac{1}{7}S=\frac{10}{7^2}+\frac{10}{7^3}+....+\frac{10}{7^{11}}\)
\(\rightarrow\)\(\left(1-\frac{1}{7}\right).S=\frac{10}{7}-\frac{10}{7^{11}}\)
=> \(S=\frac{10.7^{10}-10}{7^{10}.6}\)