\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005.2006}\)

\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(A=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)

\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)

\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)

\(3010B=\frac{1004+2006}{1004.2006}+\frac{1005+2005}{1005.2005}+...+\frac{2006+1004}{2006.1004}\) ( sửa đề nhé ) 

\(3010B=\frac{1}{2006}+\frac{1}{1004}+\frac{1}{2005}+\frac{1}{1005}+...+\frac{1}{1004}+\frac{1}{2006}\)

\(3010B=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)

\(B=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)

\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\) hay \(\frac{A}{B}\inℤ\)

Vậy ... 

Chúc bạn học tốt ~ 

Ta có:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{9.10}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)

\(\Rightarrow A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)

\(\Rightarrow A=\left(\frac{1}{6}+\frac{1}{10}\right)+\left(\frac{1}{7}+\frac{1}{9}\right)+\frac{1}{8}\)

\(\Rightarrow A=\left(\frac{10}{6.10}+\frac{6}{6.10}\right)+\left(\frac{9}{7.9}+\frac{7}{7.9}\right)+\frac{8}{8.8}\)

\(\Rightarrow A=\frac{16}{6.10}+\frac{16}{7.9}+\frac{8}{8.8}\)

\(\Rightarrow A=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)\)

Ta lại có:

\(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)

\(\Rightarrow B=\left(\frac{1}{6.10}+\frac{1}{6.10}\right)+\left(\frac{1}{7.9}+\frac{1}{7.9}\right)+\frac{1}{8.8}\)

\(\Rightarrow B=\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\)

Vậy : 

\(A:B=8\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right):\left(\frac{2}{6.10}+\frac{2}{7.9}+\frac{1}{8.8}\right)=8\)

Vậy \(A:B=8\)

\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)

\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)

Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

giúp mk vs mn ơi. mình cần gấp chiều mai nộp òi

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(\Rightarrow A< 1\)

b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)

\(\Rightarrow2B< 1\)

\(\Rightarrow B< \frac{1}{2}\)