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\(\frac{1}{5}A=\frac{1}{5}.\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{20}}\right)\)
\(\Rightarrow\frac{1}{5}A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{20}}\)
\(\Rightarrow\frac{1}{5}A-A=\left(\frac{1}{5^2}+...+\frac{1}{5^{21}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{20}}\right)\)
\(-\frac{4}{5}A=\frac{1}{5^{21}}-\frac{1}{5}\)
\(\Rightarrow A=\left(\frac{1}{5^{21}}-\frac{1}{5}\right):\left(-\frac{4}{5}\right)\)
các câu còn lại tương tự thôi
B1 c2
dùng xích ma \(\text{∑}^{20}_1\left(\frac{1}{5^x}\right)=0,25=\frac{1}{4}\)
chỗ phía dưới là 1 nha nó bị che
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
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1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
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