A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)

\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)

\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)

\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)

Câu b tương tự nha

a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)

\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)

\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)

\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)

\(\frac{A}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\)

\(A-\frac{A}{7}=\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\right)\)

\(\frac{6}{7}A=\frac{1}{7}-\frac{1}{7^{101}}\)

\(A=\left(\frac{1}{7}-\frac{1}{7^{101}}\right).\frac{7}{6}\)

\(A=\frac{1}{6}-\frac{1}{6.7^{100}}\)

\(B=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)

\(=4.\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)\)

Gọi \(C=\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\)

\(\frac{C}{5}=\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\)

\(C-\frac{5}{C}=\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\right)\)

\(\frac{4}{5}C=\frac{1}{5}-\frac{1}{5^{201}}\)

\(C=\left(\frac{1}{5}-\frac{1}{5^{201}}\right).\frac{5}{4}\)

\(=\frac{1}{4}-\frac{1}{4.5^{200}}\)

Thay vào B ta có

\(B=4.\left(\frac{1}{4}-\frac{1}{4.5^{200}}\right)\)

=\(=1-\frac{1}{5^{200}}\)

A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)

\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)

1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60. 

Thừa số phụ:

60:12 =5; 60:15=4

Ta được:

\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)

\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)

 b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252. 

Thừa số phụ:

252:7 = 36; 252:9 = 28; 252:12 = 21

Ta được:

\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)

\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)

\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)

2. a) Ta có BCNN(8, 24) = 24 nên:

\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)

 b) Ta có BCNN(12, 16) = 48 nên:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{2}{4}\)

\(=\frac{3}{4}\)

\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)

\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)

\(=\frac{25}{20}-\frac{-1}{20}\)

\(=\frac{26}{20}\)

\(=\frac{13}{10}\)

1/2+3/4-(3/4-4/5)

1/2+3/4+3/4+4/5

1/2+6/4+4/5

10/20+30/20+16/20

56/20=14/5