\(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+\frac{4}{11}\cdot\frac{4}{15}+...+\frac{4}{95}\cdot\frac{4}{99}\)

\(A=\frac{16}{3\cdot7}+\frac{16}{7\cdot11}+\frac{16}{11\cdot15}+...+\frac{16}{95\cdot99}\)

\(A=4\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{95\cdot99}\right)\)

\(A=4\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)

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

\(A=4\cdot\frac{32}{99}\)

\(A=\frac{128}{99}\)

\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{95}\times\frac{4}{99}\)

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

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

     \(=4\times\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)

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

     \(=4\times\frac{32}{99}\)

     \(=\frac{128}{99}\)

128/99

A = \(\dfrac{4}{3}\) . \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) . \(\dfrac{4}{11}\) + \(\dfrac{4}{11}\) . \(\dfrac{4}{15}\) + ... + \(\dfrac{4}{95}\) . \(\dfrac{4}{99}\)

A = \(\dfrac{4.4}{3.7}\) + \(\dfrac{4.4}{7.11}\) + \(\dfrac{4.4}{11.15}\) + ... + \(\dfrac{4.4}{95.99}\)

A = \(\dfrac{16}{3.7}\) + \(\dfrac{16}{7.11}\) + \(\dfrac{16}{11.15}\) + ... + \(\dfrac{16}{95.99}\)

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

A = 4.( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\))

A = 4.(\(\dfrac{1}{3}\) - \(\dfrac{1}{99}\))

A = 4.(\(\dfrac{33}{99}\) + \(\dfrac{-1}{99}\))

A = 4. \(\dfrac{32}{99}\)

A = \(\dfrac{4.32}{99}\)

A = \(\dfrac{128}{99}\)

Vậy A = \(\dfrac{128}{99}\)

128/99

\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{91}\times\frac{4}{95}+\frac{4}{95}\times\frac{4}{99}\)

\(=4\left(\frac{1}{3\times7}+\frac{1}{7.11}+...+\frac{1}{91\times95}+\frac{1}{95\times99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{99}\right)=4\times\frac{32}{99}=\frac{128}{99}\)

A = 4/3 x 7 + 4/7 x 11 + 4/11 x 15 + .... + 4/95 x 99

A = 4/3 - 4/7 + 4/7 - 4/11 + 4/11 - 4/15 + ..... + 4/95 - 4/99

A = 4/3 - 4/99

A = 128/99

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

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

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

\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}=\frac{128}{99}\)

\(A=\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+...+\frac{4}{95}.\frac{4}{99}\)

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

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

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

\(A=4.\frac{32}{99}\)

\(A=\frac{128}{99}\)

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

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

\(A=4.\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)

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

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

\(\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+\frac{4}{11}.\frac{4}{15}+...+\frac{4}{95}.\frac{4}{99}\)

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

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

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

sai rồi

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

\(\frac{A}{4}=\frac{7-3}{3.7}+\frac{11-7}{7.11}+...+\frac{99-95}{95.99}\)

\(\frac{A}{4}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(A=\frac{4.32}{99}\)

\(4.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{95}-\frac{1}{99}\\ 4.A=\frac{1}{3}-\frac{1}{99}\\ 4.A=\frac{32}{99}\\ A=\frac{32}{99}:4\\ A=\frac{8}{99}\)

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

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

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

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

Vậy \(A=\frac{32}{99}\)