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)

đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)

\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)

Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)

\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)

\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)

\(9B=1-\frac{1}{10^4}\)

\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)

\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)

Nhưng có vô hạn số hạng thì sao bạn

= \(\frac{5\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}{10\times\left(\frac{1}{7}+\frac{1}{3}-\frac{1}{9}\right)}\)

=\(\frac{5}{10}\)

=\(\frac{1}{2}\)

9/2

k mình nha

= \(\frac{9}{2}\)

ket qua bang 9/2

\(=\frac{1+2+3+4+5+6+7+8+9}{10}\)

\(=\frac{45}{10}=\frac{9}{2}\)

\(=\frac{45}{10}=\frac{9}{2}\)

=\(\frac{1+2+3+4+5+6+7+8+9}{10}\)

=\(\frac{45}{10}\)

=\(\frac{9}{2}\)

​

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5.31-\frac{5.2}{7}-\frac{5}{11}+\frac{5}{23}}{13.31-\frac{13.2}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{1}{3}=\frac{44}{13}\)

Bạn tham khảo nhé 

Ta có : 

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{5.31-5.\frac{2}{7}-5.\frac{1}{11}+5.\frac{1}{23}}{13.31-13.\frac{2}{7}-13.\frac{1}{11}+13.\frac{1}{23}}+\frac{3.\frac{1}{5}+3.\frac{1}{13}-3.\frac{3}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)

\(A=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{3\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)

\(A=\frac{5}{13}+\frac{3}{1}=\frac{5}{13}+\frac{39}{13}=\frac{44}{13}\)

Vậy \(A=\frac{44}{13}\)

44/13

giup mk vs cần gấp nk

\(\frac{\left(13\frac{1}{4}-2\frac{5}{7}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}=\frac{\left(\frac{53}{4}-\frac{19}{7}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{13}{10}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}\)

\(=\frac{\left(\frac{1113}{84}-\frac{228}{84}-\frac{910}{84}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{39}{30}+\frac{100}{30}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}\)

\(=\frac{\frac{-25}{84}.\frac{5751}{25}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)

\(=\frac{\frac{-1917}{28}+\frac{1309}{28}}{\frac{139}{30}.\frac{-21}{41}}\)

\(=\frac{\frac{-608}{28}}{\frac{-973}{410}}=\frac{-152}{7}.\frac{410}{-973}=\frac{62320}{6811}\)