\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(< =>2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(< =>2A-A=1-\frac{1}{2^{99}}< =>A=1-\frac{1}{2^{99}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{99}}\)

\(\Rightarrow A=1-\frac{1}{2^{99}}\)

Bài làm:

Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)

=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)

Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)

Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

có \(\frac{1}{2\cdot3}< \frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3\cdot4}< \frac{1}{3^2}< \frac{1}{2\cdot3}\)

\(\frac{1}{4\cdot5}< \frac{1}{4^2}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{9\cdot10}< \frac{1}{9^2}< \frac{1}{8\cdot9}\)

\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}>A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}\)

\(\Rightarrow\frac{8}{9}>A>\frac{2}{5}\)

Bạn ơi, sai rồi, mình k nhầm
làm sao mà \(\frac{1}{2^2}< \frac{1}{1.2}\)được

\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2018}{2017}.\frac{2019}{2018}\)

\(=\frac{3.4.5.6.....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)

\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2019}{2018}\)

\(=\frac{3.4.5....2019}{2.3.4....2018}\)

\(=\frac{2019}{2}\)

b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)

=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]

=2/-3 

=-2/3

\(=\frac{-\frac{1}{4}.-\frac{1}{5}}{\frac{5}{9}-\frac{13}{12}}\)

\(=\frac{\frac{1}{20}}{\frac{1}{4}}\)

\(=\frac{1}{20}:\frac{1}{4}\)

\(=\frac{1}{5}\)

\(\frac{-3}{14}\) - \(\frac{5}{-14}\)

= \(\frac{-3}{14}+\frac{5}{14}\)

= \(\frac{2}{14}\)

= \(\frac17\)

- \(\frac54\) - \(\frac34\)

= -(\(\frac54+\frac34\))

= - \(\frac84\)

= - 2

\(\frac{15}{6}-\frac{-10}{20}\)

= \(\frac52\) + \(\frac12\)

= \(\frac62\)

= 3

\(\frac{26}{-35}\) - \(\frac{6}{35}\)

= - (\(\frac{26}{35}\) + \(\frac{6}{35}\))

=- \(\frac{32}{35}\)