A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

A = \(\frac{1}{3}-\frac{1}{99}\)

A = \(\frac{32}{99}\)

\(\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+...+\frac{2}{95.99}\)

\(=\frac{1}{2}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{98}{99}\)

\(=\frac{49}{99}\)

Chúc cậu học tốt !!!

\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{95.99}\)

\(A=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\frac{98}{99}\)

\(A=\frac{49}{198}\)

Giờ ta so sánh : 

\(A=\frac{49}{198}\) và B=1 

Ta thấy :

\(\frac{49}{198}<1\)

=> A < B

Vậy A < B

\(A=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

S=2/3.7+2/7.1+2/11.15+...+2/95.99

S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}\right)+...+\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{99}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

S = \(\frac{1}{2}.\frac{32}{99}\)

S = \(\frac{16}{99}\)

M=1/2+1/6+1/12+1/20+..+1/110

M = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)

M = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

M = \(1-\frac{1}{11}\)

M = \(\frac{10}{11}\)

\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

\(B=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

Vậy giá trị của biểu thức \(B=\frac{32}{99}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

Ta Có

(1/3-1/7+1/7-1/11+1/11-1/15+...+1/95-1/99)

(1/3-1/99)

32/99

A=1/3*7+1/7*11+..+1/95*99

=> 4A=4/3*7+4/7*11+..+4/95*99

=>4A=1/3-1/7+1/7-1/11+...+1/95-1/99=1/3-1/99=32/99

=>A=8/99

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\cdot\frac{32}{99}=\frac{8}{99}\)

Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)

\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)

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

\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)

\(=2-\frac{1}{5}\)< 2 

Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2