\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)

\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)

\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)

\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)

\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)

\(B=\dfrac{4.9.16.100}{3.8.15.99}\)

\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)

\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)

\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)

2

a. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

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

=\(\dfrac{1}{2}-\dfrac{1}{100}\)

=\(\dfrac{49}{100}\)

Ta có \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2-1+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).

Từ đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\); \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\); \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\); \(1+\dfrac{1}{4.6}=\dfrac{5^2}{4.6}\);...; \(1+\dfrac{1}{2022.2024}=\dfrac{2023^2}{2022.2024}\).

Suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2022.2024}\right)\)

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{2023^2}{2022.2024}\)

\(=\dfrac{2.2023}{2024}\) \(=\dfrac{2023}{1012}\)

A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)

\(=\dfrac{5}{22}\)

Mk có thể làm wen đc k

=1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100

=1/2 - 1/100 = 49/100

1/2 - 1/4 +  1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100

= 1/2 - 1/100 

= 49/100

1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100

= 1/2 . ( 2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100 )

= 1/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100 )

= 1/2 . ( 1/2 - 1/100 )

= 1/2 . ( 50/100 - 1/100 )

= 49/200

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)

Gọi biểu thức trên là A

Ta có:

2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2

2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)

2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)

2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{x+2}\)

mà A = \(\dfrac{1}{10}\)(đề bài)

nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)

                     suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\) 

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)

\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)