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

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)

\(A=\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\)

\(10A=7+\frac{7}{10}+...+\frac{7}{10^{99}}\)

\(\Rightarrow10A-A=9A=7-\frac{7}{10^{100}}\)

Ta có : \(10A=7+\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{99}}\)

                               \(A=\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{99}}+\frac{7}{10^{100}}\)

       \(\Rightarrow9A=10A-A=7-\frac{7}{10^{100}}\)

        \(\Rightarrow A=\frac{7-\frac{7}{10^{100}}}{9}\)

FGHFFGGDJJG

giup mk vs cần gấp nk

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

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

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

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

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

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 \(A< 3\)

b) \(A=\frac{1+5+5^2+5^3+...+5^{10}+5^{11}}{1+5+5^2+5^3+...+5^9+5^{10}}=5^{11}\)

bn rút gọn là dc  

\(B=\frac{1+7+7^2+7^3+...+7^{10}+7^{11}}{1+7+7^2+7^3+...+7^9+7^{10}}=7^{11}\)

\(A=5^{11},B=7^{11}\)

\(\Rightarrow7^{11}>5^{11}\Rightarrow B>A\)

hk tốt #